MCQ 11 Mark
A particle executing SHM of amplitude $5 cm$ has an acceleration of $27 cm / s ^2$ when it is $3 cm$ from the mean position. Its maximum velocity is
- ✓
$15 cm / s$
- B
$30 cm / s$
- C
$45 cm / s$
- D
$60 cm / s$.
AnswerCorrect option: A. $15 cm / s$
$15 cm / s$
View full question & answer→MCQ 21 Mark
The period of SHM of a particle with maximum velocity $50 cm / s$ and maximum acceleration $10 cm / s ^2$ is
- A
$31.42 s$
- B
$6.284 s$
- ✓
$3.142 s$
- D
$0.3142 s$.
AnswerCorrect option: C. $3.142 s$
$3.142 s$
View full question & answer→MCQ 31 Mark
A simple harmonic oscillator has an amplitude $A$ and period $T$. The time required by the oscillator to cover the distance from $x = A$ to $x =\frac{A}{2}$ is
- A
$\frac{T}{2}$
- B
$\frac{T}{3}$
- C
$\frac{T}{4}$
- ✓
$\frac{T}{6}$
AnswerCorrect option: D. $\frac{T}{6}$
$\frac{T}{6}$
View full question & answer→MCQ 41 Mark
A vertical spring-and-block system has a block of mass $10 g$ and oscillates with a period $1 s$. The period of SHM of a block of mass $90 g$, suspended from the same spring, is
- A
$\frac{1}{9} s$
- B
$\frac{1}{3} s$
- ✓
$3 s$
- D
$9 s$.
View full question & answer→MCQ 51 Mark
A horizontal spring-and-block system consists of a block of mass $1 kg$, resting on a frictionless surface, and an ideal spring. A force of $10 N$ is required to compress the spring by $10 cm$. The spring constant of the spring is
- A
$100 N \cdot m ^{-1}$
- B
$10 N \cdot m ^{-1}$
- ✓
$N \cdot m ^{-1}$
- D
$0.1 N \cdot m ^{-1}$.
AnswerCorrect option: C. $N \cdot m ^{-1}$
$N \cdot m ^{-1}$
View full question & answer→MCQ 61 Mark
A spring-and-block system constitutes a simple harmonic oscillator. To double the frequency of oscillation, the mass of the block must be the initial mass.
AnswerCorrect option: A. $\frac{1}{4}$ times
$\frac{1}{4}$ times
View full question & answer→MCQ 71 Mark
A magnet is suspended to oscillate in the horizontal plane. It makes 20 oscillations per minute at a place where the dip angle is $30^{\circ}$ and 15 oscillations per minute where the dip angle is $60^{\circ}$. The ratio of the Earth's total magnetic field at the two places is
- A
$3 \sqrt{3}: 16$
- ✓
$16: 9 \sqrt{3}$
- C
$4: 9 \sqrt{3}$
- D
$9: 16 \sqrt{3}$.
AnswerCorrect option: B. $16: 9 \sqrt{3}$
$16: 9 \sqrt{3}$
View full question & answer→MCQ 81 Mark
Two bar magnets of identical size have magnetic moments $M_A$ and $M_B$. If the magnet $A$ oscillates at twice the frequency of magnet $B$, then
- A
$M_A=2 M_B$
- B
$M_A=8 M_B$
- ✓
$M_A=4 M_B$
- D
$M_B=8 M_A$.
AnswerCorrect option: C. $M_A=4 M_B$
$M_A=4 M_B$
View full question & answer→MCQ 91 Mark
If the equation of motion of a particle performing SHM is $x=0.028 \cos (2.8 \pi t+\pi)$ (all quantities in SI units), the frequency of the motion is
- A
$0.7 Hz$
- ✓
$1.4 Hz$
- C
$2.8 Hz$
- D
$14 Hz$.
AnswerCorrect option: B. $1.4 Hz$
$1.4 Hz$
View full question & answer→MCQ 101 Mark
Two springs of force constants $k_1$ and $k_2\left(k_1>k_2\right)$ are stretched by the same force. If $W_1$ and $W_2$ be the work done in stretching the springs, then
- A
$W_1=W_2$
- ✓
$W_1<W_2$
- C
$W _1> W _2$
- D
$W _1= W _2=0$.
AnswerCorrect option: B. $W_1<W_2$
$W_1<W_2$
View full question & answer→MCQ 111 Mark
The average displacement over a period of SHM is (A = amplitude of SHM)
MCQ 121 Mark
Two particles perform linear simple harmonic motion along the same path of length 2A and period T as shown in the graph below. The phase difference between them is:
- A
- ✓
$\frac{\pi}{4} rad$
- C
$\frac{\pi}{2} rad$
- D
$\frac{3 \pi}{4}$ rad
AnswerCorrect option: B. $\frac{\pi}{4} rad$
$\frac{\pi}{4} rad$
View full question & answer→MCQ 131 Mark
The total work done by a restoring force in simple harmonic motion of amplitude $A$ and angular frequency $\omega$, in one oscillation is
View full question & answer→MCQ 141 Mark
A seconds pendulum is suspended in an elevator moving with a constant speed in the downward direction. The periodic time (T) of that pendulum is
- A
- ✓
- C
- D
very much greater than two seconds.
View full question & answer→MCQ 151 Mark
If $T$ is the time period of a simple pendulum in an elevator at rest, its time period in a freely falling elevator will be
- A
$\frac{T}{\sqrt{2}}$
- B
$\sqrt{2} T$
- C
$2 T$
- ✓
View full question & answer→MCQ 161 Mark
In 20 s, two simple pendulums, $P$ and $Q$, complete 9 and 7 oscillations, respectively, on the Earth. On theMoon, where the acceleration due to gravity is $\frac{1}{6}$ th that on the Earth, their periods are in the ratio
- A
$8: 1$
- B
$9: 7$
- ✓
$7: 9$
- D
$3: 14$.
AnswerCorrect option: C. $7: 9$
(C) $7: 9$
View full question & answer→MCQ 171 Mark
The amplitude of oscillations of a simple pendulum of period $T$ and length $L$ is increased by $5 \%$. The new period of the pendulum will be
- A
$T / 8$
- B
$T / 4$
- C
$T / 2$
- ✓
$T$.
View full question & answer→MCQ 181 Mark
If the length of a simple pendulum is doubled keeping its amplitude constant, its energy will be
- A
- B
- ✓
- D
increased to four times the initial energy.
View full question & answer→MCQ 191 Mark
If the length of a simple pendulum is increased to 4 times its initial length, its frequency of oscillation will
AnswerCorrect option: A. reduce to half its initial frequency
reduce to half its initial frequency
View full question & answer→MCQ 201 Mark
The phase change of a particle performing SHM between successive passages through the mean position is
- A
$2 \pi rad$
- ✓
$\pi rad$
- C
$\frac{\pi}{2} rad$
- D
$\frac{\pi}{4} rad$.
AnswerCorrect option: B. $\pi rad$
$\pi rad$
View full question & answer→MCQ 211 Mark
Two spring-and-block oscillators oscillate harmonically with the same amplitude and a constant phase difference of $90^{\circ}$. Their maximum velocities are $v$ and $v+x$. The value of $x$ is
- ✓
- B
$\frac{v}{3}$
- C
- D
$\frac{v}{\sqrt{2}}$.
View full question & answer→MCQ 221 Mark
The total energy of a particle executing SHM is proportional to
- A
the frequency of oscillation
- ✓
the square of the amplitude of motion
- C
the velocity at the equilibrium position
- D
the displacement from the equilibrium position.
AnswerCorrect option: B. the square of the amplitude of motion
the square of the amplitude of motion
View full question & answer→MCQ 231 Mark
A particle performing linear SHM with a frequency $n$ is confined within limits $x= \pm A$. Midway between an extremity and the equilibrium position, its speed is
- A
$\sqrt{6} nA$
- ✓
$\sqrt{3} \pi n A$
- C
$\sqrt{6} \pi AA$
- D
$\sqrt{12} \pi nA$
AnswerCorrect option: B. $\sqrt{3} \pi n A$
$\sqrt{3} \pi n A$
View full question & answer→MCQ 241 Mark
The acceleration of a particle performing SHM is $3 m / s ^2$ at a distance of $3 cm$ from the mean position.The periodic time of the motion is
- A
$0.02 \pi s$
- B
$0.04 \pi s$
- ✓
$0.2 \pi s$
- D
$2 \pi s$.
AnswerCorrect option: C. $0.2 \pi s$
$0.2 \pi s$
View full question & answer→MCQ 251 Mark
In simple harmonic motion, the acceleration of a particle is zero when its
- A
- ✓
- C
both velocity and displacement are zero
- D
both velocity and displacement are maximum.
View full question & answer→MCQ 261 Mark
The minimum time taken by a particle in SHM with period T to go from an extreme position to a point half way to the equilibrium position is
- A
$\frac{T}{12}$
- B
$\frac{T}{8}$
- ✓
$\frac{T}{6}$
- D
$\frac{T}{4}$
AnswerCorrect option: C. $\frac{T}{6}$
$\frac{T}{6}$
View full question & answer→MCQ 271 Mark
A particle executes linear SHM with period $12 s$. To traverse a distance equal to half its amplitude from the equilibrium position, it takes
- A
$6 s$
- B
$4 s$
- C
$2 s$
- ✓
$1 s$.
AnswerCorrect option: D. $1 s$.
$1 s$
View full question & answer→MCQ 281 Mark
A particle executing linear SHM has velocities $v _1$ and $v _2$ at distances $x _1$ and $x _2$, respectively, from the mean position. The angular velocity of the particle is
- A
$\sqrt{\frac{x_1^2-x_2^2}{v_2^2-v_1^2}}$
- ✓
$\sqrt{\frac{v_2^2-v_1^2}{x_1^2-x_2^2}}$
- C
$\sqrt{\frac{x_1^2+x_2^2}{v_2^2+v_1^2}}$
- D
$\sqrt{\frac{v_2^2+v_1^2}{x_1^2+x_2^2}}$.
AnswerCorrect option: B. $\sqrt{\frac{v_2^2-v_1^2}{x_1^2-x_2^2}}$
$\sqrt{\frac{v_2^2-v_1^2}{x_1^2-x_2^2}}$
View full question & answer→MCQ 291 Mark
A spring-and-block oscillator with an ideal spring of force constant $180 N / m$ oscillates with a frequency of $6 Hz$. The mass of the block is, approximately,
- ✓
$\frac{1}{8} kg$
- B
$\frac{1}{4} kg$
- C
$4 kg$
- D
$8 kg$.
AnswerCorrect option: A. $\frac{1}{8} kg$
$\frac{1}{8} kg$
View full question & answer→MCQ 301 Mark
A particle performs linear SHM with a period of $6 s$, starting from the positive extremity. At time $t=7 s$, its displacement is $3 cm$. The amplitude of the motion is
- A
$4 cm$
- ✓
$6 cm$
- C
$8 cm$
- D
$12 cm$.
AnswerCorrect option: B. $6 cm$
$6 cm$
View full question & answer→MCQ 311 Mark
The differential equation of SHM for a seconds pendulum is
- A
$\frac{d^2 x}{d t^2}+ x =0$
- B
$\frac{d^2 x}{d t^2}+\pi x=0$
- C
$\frac{d^2 x}{d t^2}+4 \pi x=0$
- ✓
$\frac{d^2 x}{d t^2}+\pi^2 x=0$.
AnswerCorrect option: D. $\frac{d^2 x}{d t^2}+\pi^2 x=0$.
$\frac{d^2 x}{d t^2}+\pi^2 x=0$.
View full question & answer→MCQ 321 Mark
The graph shows variation of displacement of a particle performing S.H.M. with time t. Which of the following statements is correct from the graph?
- A
The acceleration is maximum at time T.
- ✓
The force is maximum at time 3T/4.
- C
The velocity is zero at time T/2.
- D
The kinetic energy is equal to total energy at time T/4.
AnswerCorrect option: B. The force is maximum at time 3T/4.
The force is maximum at time 3T/4.
View full question & answer→MCQ 331 Mark
Two identical springs of constant k are connected, first in series and then in parallel. A metal block of mass m is suspended from their combination. The ratio of their frequencies of vertical oscillations will be in a ratio
- A
$ 1: 4 $
- ✓
$ 1: 2 $
- C
$ 2: 1 $
- D
$ 4: 1 $
AnswerCorrect option: B. $ 1: 2 $
$ 1: 2 $
View full question & answer→MCQ 341 Mark
The length of second’s pendulum on the surface of earth is nearly 1 m. Its length on the surface of moon should be [Given: acceleration due to gravity (g) on moon is 1/6 th of that on the earth’s surface]
- ✓
$\frac{1}{6} m$
- B
$6 m$
- C
$\frac{1}{36} m$
- D
$\frac{1}{\sqrt{6}} m$.
AnswerCorrect option: A. $\frac{1}{6} m$
$\frac{1}{6} m$
View full question & answer→MCQ 351 Mark
A body of mass 1 kg is performing linear S.H.M. Its displacement x (cm) at t (second) is given by x = 6 sin (100t + π/4). Maximum kinetic energy of the body is
MCQ 361 Mark
A particle performs linear S.H.M. starting from the mean position. Its amplitude is A and time period is T. At the instance when its speed is half the maximum speed, its displacement x is
- ✓
$\frac{\sqrt{3}}{2} A$
- B
$\frac{2}{\sqrt{3}}$
- C
$A / 2$
- D
$\frac{1}{\sqrt{2}}$
AnswerCorrect option: A. $\frac{\sqrt{3}}{2} A$
$\frac{\sqrt{3}}{2} A$
View full question & answer→MCQ 371 Mark
Equation of simple harmonic progressive wave is given by $y=\frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t$, then the resultant amplitude of the wave is $\left(\cos 90^{\circ}=0\right)$
AnswerCorrect option: D. $\sqrt{\frac{a+b}{a b}}$
(d) : $y=\frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t$ $=\frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \sin \left(\omega t+\frac{\pi}{2}\right)$
Phase difference $=\frac{\pi}{2}$
Resultant amplitude
$
=\sqrt{\left(\frac{1}{\sqrt{a}}\right)^2+\left(\frac{1}{\sqrt{b}}\right)^2}=\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{\frac{a+b}{a b}}
$
View full question & answer→MCQ 381 Mark
A simple pendulum of length $2 m$ is given a horizontal push through angular displacement of $60^{\circ}$. If the mass of bob is $200$ gram, the angular velocity of the bob will be $($Take acceleration due to gravity $=10\ m / s ^2)$
$\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2\right)$
- A
$2 \sqrt{2} \ rad / s$
- B
$3 \sqrt{2} \ rad / s$
- ✓
$2 \sqrt{2.5} \ rad / s$
- D
$3 \sqrt{2.5} \ rad / s$
AnswerCorrect option: C. $2 \sqrt{2.5} \ rad / s$
Given$, \theta=60^{\circ},$
$L=2 m , A B=L \cos \theta$
$h=L-L \cos \theta=L(1-\cos \theta)$
$\text { P.E. }=\text { K.E. }$
$m g L(1-\cos \theta)=\frac{1}{2} m v^2$
$2 \times 10 \times 2\left(1-\frac{1}{2}\right)=\frac{1}{2} v^2$
$v^2=4 \times 10 $
$\Rightarrow v=2 \sqrt{10} m / s$
$\omega=\frac{v}{L}=\frac{2 \sqrt{10}}{2}$
$=2 \sqrt{2.5}\ rad / s $
View full question & answer→MCQ 391 Mark
A particle is vibrating in S.H.M. with an amplitude of $4 cm$. At what displacement from the equilibrium position is its energy half potential and half kinetic?
- A
$1 cm$
- B
$\sqrt{2} cm$
- C
$2 cm$
- ✓
$2 \sqrt{2} cm$
AnswerCorrect option: D. $2 \sqrt{2} cm$
(d) : Given, $A=4 cm$
$
\begin{aligned}
& \frac{ PE }{2}=\frac{ KE }{2} ; \frac{1}{2} \times \frac{1}{2} m \omega^2 y^2=\frac{1}{2} \times \frac{1}{2} \times m \omega^2\left(A^2-y^2\right) \\
& y^2=A^2-y^2 ; 2 y^2=A^2 \Rightarrow y=\frac{A}{\sqrt{2}}=2 \sqrt{2} cm
\end{aligned}
$
View full question & answer→MCQ 401 Mark
A spring has a certain mass suspended from it and its period for vertical oscillations is ' $T_1$ '. The spring is now cut into two equal halves and the same mass is suspended from one of the halves. The period of vertical oscillations is now ' $T_2$ ' The ratio $T_1 / T_2$ is
- A
- ✓
$\sqrt{2}$
- C
$\frac{1}{\sqrt{2}}$
- D
$\frac{1}{2}$
AnswerCorrect option: B. $\sqrt{2}$
(b) : When the spring cut into half, the spring constant becomes doubled.
As, $T \propto \frac{1}{\sqrt{k}} \quad \therefore \frac{T_2}{T_1}=\sqrt{\frac{k_1}{k_2}}=\sqrt{\frac{k}{2 k}} \quad \Rightarrow \frac{T_1}{T_2}=\sqrt{2}$
View full question & answer→MCQ 411 Mark
Two S.H.Ms are represented by equations $y_1=0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)$ and $y_2=0.1 \cos (100 \pi t)$ The phase difference between the speeds of the two particles is
- A
$\frac{\pi}{3}$
- ✓
$-\frac{\pi}{6}$
- C
$+\frac{\pi}{6}$
- D
$-\frac{\pi}{3}$
AnswerCorrect option: B. $-\frac{\pi}{6}$
(b) : $y_1=0.1 \sin (100 \pi t+\pi / 3), y_2=0.1 \cos (100 \pi t)$
$
\begin{aligned}
& v_1=\frac{d y_1}{d t}=0.1 \times 100 \pi \cos \left(100 \pi t+\frac{\pi}{3}\right) ...(i)\\
& v_2=\frac{d y_2}{d t}=0.1 \times 100 \pi \cos \left(100 \pi t+\frac{\pi}{2}\right)...(ii)
\end{aligned}
$
From (i) and (ii), $\Delta \phi=\frac{\pi}{3}-\frac{\pi}{2}=-\frac{\pi}{6}$
View full question & answer→MCQ 421 Mark
A seconds pendulum is placed in a space laboratory orbiting round the earth at a height ' $3 R$ ' from the earth's surface. The time period of the pendulum will be ( $R=$ radius of earth)
Answer(d) : In laboratory $g_{\text {eff }}=0$
$
\therefore \quad T=2 \pi \sqrt{\frac{L}{g_{\text {eff }}}}=2 \pi \sqrt{\frac{L}{0}}=\infty
$
View full question & answer→MCQ 431 Mark
A rubber ball filled with water, having a small hole is used as the bob of a simple pendulum. The time period of such a pendulum
- A
- B
- C
- ✓
first increases and then decreases, finally having same value as at the beginning.
AnswerCorrect option: D. first increases and then decreases, finally having same value as at the beginning.
(d) : As the water starts leaking, the effective length increases, and thus the time period increases $(T \propto \sqrt{l})$. Now as all the water has drained off, the effective length become same as beginning. So the time period becomes same as the beginning.
View full question & answer→MCQ 441 Mark
The maximum velocity of a particle performing S.H.M. is 'V'. If the periodic time is made $\left(\frac{1}{3}\right)^{ rd }$ and the amplitude is doubled, then the new maximum velocity of the particle will be
- A
$\frac{V}{6}$
- B
$\frac{3 V }{2}$
- C
$3 V$
- ✓
$6 V$
Answer(d) : Maximum velocity, $V=A \omega$...(i)
$
T^{\prime}=\frac{T}{3} ; A^{\prime}=2 A
$
$
\omega^{\prime}=\frac{2 \pi}{T^{\prime}}=\frac{2 \pi \times 3}{T}
$
So,
$
\begin{aligned}
& V^{\prime}=A^{\prime} \omega^{\prime}=\frac{2 A \times 2 \pi \times 3}{T} \\
& V^{\prime}=6 A \omega=6 V \quad \text { (from (i)) }
\end{aligned}
$
View full question & answer→MCQ 451 Mark
A light spring is suspended with mass $m_1$ at its lower end and its upper end fixed to a rigid support. The mass is pulled down a short distance and then released. The period of oscillation is $T$ second. When a mass $m_2$ is added to $m_1$ and the system is made to oscillate, the period is found to be $\frac{3}{2} T$. The ratio of $m_1: m_2$ is
- A
$2: 3$
- B
$3: 4$
- ✓
$4: 5$
- D
$5: 6$
AnswerCorrect option: C. $4: 5$
As, $T=2 \pi \sqrt{\frac{m}{K}}$$T=2 \pi \sqrt{\frac{m_1}{K}} \ldots \text { (i) } ;$
$\frac{3 T}{2}=2 \pi \sqrt{\frac{m_1+m_2}{K}} \ldots \text { (ii) }$
Divide equation $(i)$ by equation $(ii)$
$ \frac{2}{3}=\sqrt{\frac{m_1}{m_1+m_2}} ;$
$\frac{4}{9}=\frac{m_1}{m_1+m_2}$
$4 _1+4 m_2=9 m_1$
$4 m_2=5 m_1$
$m_1: m_2=4: 5$
View full question & answer→MCQ 461 Mark
A block of mass $'M\ '$ rests on a piston executing $\text{S.H.M.}$ of period one second. The amplitude of oscillations, so that the mass is separated from the piston, is $($acceleration due to gravity, $g=10 ms ^{-2}, \pi^2=10)$
- ✓
$0.25 m$
- B
$0.5 m$
- C
$1 m$
- D
$\infty$
AnswerCorrect option: A. $0.25 m$
$T=1 s , g=10 m / s ^2$
$T=2 \pi \sqrt{\frac{m}{K}}$
For no contact of block and spring
Spring force $\geq$ Weight $K x \geq m g$
$x \geq \frac{m g}{K} $
$\Rightarrow x \geq \frac{T^2}{4 \pi^2} g\ ($ As $\omega=\sqrt{\frac{k}{m}})$
$x \geq \frac{1 \times 1 \times 10}{4 \times 10}$
$x=0.25 m$
View full question & answer→MCQ 471 Mark
- A
$230 N m ^{-1}$
- B
$120 N m ^{-1}$
- ✓
$60 N m ^{-1}$
- D
$30 N m ^{-1}$
AnswerCorrect option: C. $60 N m ^{-1}$
(c) : The springs are in parallel combination.
$
K_{\text {eff }}=2 K
$
Time period is given by $T=2 \pi \sqrt{\frac{M}{K_{\text {eff }}}}$
$
\begin{aligned}
& 2=2 \pi \sqrt{\frac{12}{2 K}} ; 1=\pi^2 \times \frac{12}{2 K} ; 2 K=120 \quad\left(\therefore \pi^2=10\right) \\
& K=60 N / m
\end{aligned}
$
View full question & answer→MCQ 481 Mark
The displacement of a particle executing S.H.M. is $x=a \sin (\omega t-\phi)$. Velocity of the particle at time $t=\frac{\phi}{\omega}$ is $\left(\cos 0^{\circ}=1\right)$
Answer(b) : $x=a \sin (\omega t-\phi)$
$
\begin{aligned}
& v=\frac{d x}{d t}=a \omega \cos (\omega t-\phi) \quad At _3 t=\frac{\phi}{\omega} \\
& v=a \omega \cos \left(\frac{\omega \phi}{\omega}-\phi\right)=a \omega
\end{aligned}
$
View full question & answer→MCQ 491 Mark
The bob of simple pendulum of length $L$ is released from a position of small angular displacement $\theta$. Its linear displacement at time $t$ is ( $g=$ acceleration due to gravity)
- ✓
$L \theta \cos \left[\sqrt{\frac{g}{L}} \cdot t\right]$
- B
$L \theta \sin \left[2 \pi \sqrt{\frac{g}{L}} \cdot t\right]$
- C
$L \theta \cos \left[2 \pi \sqrt{\frac{g}{L}} \cdot t\right]$
- D
$L \theta \sin \left[\sqrt{\frac{g}{L}} \cdot t\right]$
AnswerCorrect option: A. $L \theta \cos \left[\sqrt{\frac{g}{L}} \cdot t\right]$
(a) : As, $\theta=\theta_0 \cos \omega t$ and $\omega=\frac{2 \pi}{T}$
$
T=2 \pi \sqrt{\frac{L}{g}} ; \omega=\sqrt{\frac{g}{L}}
$
Angular displacement, $\theta^{\prime}=\theta \cos \sqrt{\frac{g}{L}} t$
Linear displacement, $x=L \theta^{\prime}$
$
\therefore \quad x=L \theta \cos \sqrt{\frac{g}{L}} t
$
View full question & answer→MCQ 501 Mark
A simple pendulum of length ' $l$ ' and a bob of mass ' $m$ ' is executing S.H.M. of small amplitude ' $A$ '. The maximum tension in the string will be ( $g=$ acceleration due to gravity)
- A
$2 m g$
- ✓
$m g\left[1+\left(\frac{A}{l}\right)^2\right]$
- C
$m g\left[1+\left(\frac{A}{l}\right)\right]^2$
- D
$m g\left[1+\left(\frac{A}{l}\right)\right]$
AnswerCorrect option: B. $m g\left[1+\left(\frac{A}{l}\right)^2\right]$
(b) : Tension in the string will be maximum when bob will be passing through mean position.
$
\therefore T=m g \cos \theta+\frac{m v^2}{l}
$
$T$ is maximum when, $\cos \theta=1_l$
$
\therefore \quad T_m=m g+\frac{m v^2}{l}
$
At mean position, $\omega=\sqrt{\frac{g}{l}}$
And, $v=\omega \sqrt{A^2-x^2}$, At mean position,
$
v_{\max }=\omega A \text { or } v=\sqrt{\frac{g}{l}} A \text { or } v^2=\frac{A^2 g}{l}
$
From eq (i), we have
$
T_m=m g+\frac{m g A^2}{l^2}=m g\left[1+\frac{A^2}{l^2}\right]
$
View full question & answer→MCQ 511 Mark
The ratio of frequencies of two oscillating pendulums are $3: 2$. Their lengths are in the ratio
- A
$\sqrt{3}: \sqrt{2}$
- B
$9: 4$
- ✓
$4: 9$
- D
$\sqrt{2}: \sqrt{3}$
AnswerCorrect option: C. $4: 9$
(c) : As $T=2 \pi \sqrt{\frac{l}{g}}$ $v \propto \frac{1}{T}$ so, $v=\frac{1}{T}=\frac{1}{2 \pi} \sqrt{\frac{g}{l}}$
For pendulum $1, v_1=\frac{1}{2 \pi} \sqrt{\frac{g}{l_1}}$
For pendulum $2, v_2=\frac{1}{2 \pi} \sqrt{\frac{g}{l_2}}$
$
\frac{v_1}{v_2}=\sqrt{\frac{l_2}{l_1}} \Rightarrow \frac{3}{2}=\sqrt{\frac{l_2}{l_1}}
$
On squaring both sides, we have
$
\frac{9}{4}=\frac{l_2}{l_1} ; l_1: l_2=4: 9
$
View full question & answer→MCQ 521 Mark
A rectangular block of mass ' $M$ ' and cross-sectional area ' $A$ ' floats on a liquid of density ' $\rho$ '. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency ' $n$ ' then
- A
$n \propto A$
- B
$n \propto A^2$
- ✓
$n \propto \sqrt{A}$
- D
$n \propto A^3$
AnswerCorrect option: C. $n \propto \sqrt{A}$
(c) : Time period of SHM is given by
$
T=2 \pi \sqrt{\frac{l}{g}}...(i)
$
(where $l$ is length of rectangular block)
According to law of floatation,
Weight of block $=$ weight of the displaced liquid
$
\Rightarrow M g=A l \rho g \Rightarrow l=\frac{M}{A \rho}
$
using value of $T$ ' in equation (i), we get
$
T=2 \pi \sqrt{\frac{M}{A \rho g}}
$
As $T \propto \frac{1}{n}$ so $n \propto \sqrt{A}$.
Here ' $n$ ' is frequency of oscillations.
View full question & answer→MCQ 531 Mark
The equation of motion of a particle performing linear S.H.M is $x=5 \sin \left[4 t-\frac{\pi}{6}\right]$, where ' $x$ ' is its displacement in $cm$. The velocity of the particle when its displacement is $3 cm$, is
- A
$10 cm / s$
- ✓
$16 cm / s$
- C
$6 cm / s$
- D
$8 cm / s$
AnswerCorrect option: B. $16 cm / s$
(b) : The given equation is, $x=5 \sin \left(4 t-\frac{\pi}{6}\right)$ Amplitude, $a=5$
Angular velocity, $\omega=4$
The velocity of the wave obtained by,
$
v=\omega \sqrt{a^2-y^2}=4 \sqrt{(5)^2-(3)^2}=16 cm
$
View full question & answer→MCQ 541 Mark
Consider two SHMs along the same straight line $x_1=A_1 \sin \left(\omega t+\phi_1\right), x_2=A_2 \sin \left(\omega t+\phi_2\right)$, where $A_1$ and $A_2$ are their amplitudes and $\phi_1$ and $\phi_2$ are their initial phase angle. If the two SHMs meet simultaneously and ' $R$ ' is the resultant amplitude, match column I with column II. | Column-I | Column-II |
| A. | The two SHMs are in phase, $A_1=A_2=A$ | I. | $R=A_1+A_2$ |
| B. | The two SHMs are in phase, $A_1 \neq A_2$ | II. | $R=0$ |
| C. | The two SHMs are $90^{\circ}$ out of phase, $A_1=A_2=A$ | III. | $R=2 A$ |
| D. | The two SHMs are $180^{\circ}$out of phase, $A_1=A_2$ | IV. | $R=\sqrt{2} A$ |
Answer(a) : $x_1=A_1 \sin \left(\omega t+\phi_1\right)$
$
x_2=A_2 \sin \left(\omega t+\phi_2\right)
$
where $A_1, A_2$ are amplitudes of two SHM's and $\phi_1, \phi_2$ are phase angle.
Resultant amplitude,
$
R=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \left(\phi_1-\phi_2\right)}...(i)
$
(i) When two SHM's are in phase, $A_1=A_2=A$, then from equation (i)
$
R=\sqrt{A^2+A^2+2 A^2}=2 A
$
(ii) When two SHM's are in phase but $A_1 \neq A_2$, then
$
\begin{aligned}
& R=\sqrt{A_1^2+A_2^2+2 A_1 A_2}=\sqrt{\left(A_1+A_2\right)^2} \\
& R=A_1+A_2
\end{aligned}
$
(iii) When to SHMs are $90^{\circ}$ out of phase, then
$
R=\sqrt{A^2+A^2+0}=\sqrt{2} A
$
(iv) When two SHMs are $180^{\circ}$ out of phase and $A_1=A_2$,
$
\begin{aligned}
& R=\sqrt{A_1^2+A_1^2-2 A_1^2} \\
& R=0
\end{aligned}
$
Hence, option (a) is correct answer.
View full question & answer→MCQ 551 Mark
A particle executes SHM of type $x=A \sin \omega t$. It takes time $t_1$ from $x=0$ to $x=\frac{A}{2}$ and $t_2$ from $x=\frac{A}{2}$ to $x=A$. The ratio $t_1 ; t_2$ will be
- A
$1: 1$
- ✓
$1: 2$
- C
$1: 3$
- D
$2: 1$
AnswerCorrect option: B. $1: 2$
(b) : $t_1+t_2=\frac{T}{4}$ or $t_2=\frac{T}{4}-t_1$
At time, $t=t_1, x=\frac{A}{2}$
$\therefore \frac{A}{2}=A \sin \omega t_1$ or $\omega t_1=\frac{\pi}{6}$ or $t_1=\frac{\pi}{6 \omega}$
$\therefore \quad t_2=\frac{T}{4}-\frac{\pi}{6 \omega}=\frac{2 \pi}{4 \omega}-\frac{\pi}{6 \omega}=\frac{2 \pi}{6 \omega} \quad\left(\because T=\frac{2 \pi}{\omega}\right)$
$\therefore \quad t_1: t_2=1: 2$
View full question & answer→MCQ 561 Mark
- A
$\frac{k_1 A}{k_2}$
- B
$\frac{k_2 A}{k_1}$
- C
$\frac{k_1 A}{k_1+k_2}$
- ✓
$\frac{k_2 A}{k_1+k_2}$
AnswerCorrect option: D. $\frac{k_2 A}{k_1+k_2}$
(d) :
Here the two springs are in series. Let the mass $M$ be pulled through small distance $A$ towards right hand side. Due to it, let $x_1$ and $x_2$ be the extension in two springs. When two springs are in series, the tension or the force is same for both the springs. So $k_1 x_1=k_2 x_2$ or $x_2=\frac{k_1 x_1}{k_2}$ Total extension, $A=x_1+x_2=x_1+\frac{k_1 x_1}{k_2}$;
$
A=x_1\left(\frac{k_2+k_1}{k_2}\right)
$
$\therefore \quad$ Amplitude of point $P=$ extension of spring $k_1$
$
=x_1=\frac{k_2 A}{k_1+k_2}
$
View full question & answer→MCQ 571 Mark
If $x, V$ and $a$ denote the displacement, velocity and acceleration of a particle respectively executing simple harmonic motion of periodic time $T$, then which one of the following does not change with time?
- ✓
$\frac{a T}{x}$
- B
$a T+2 \pi V$
- C
$\frac{a T}{V}$
- D
$a T+4 \pi^2 V^2$
AnswerCorrect option: A. $\frac{a T}{x}$
(a) : Given $x, V$ and $a$ denote the displacement, velocity and acceleration of a particle respectively executing simple harmonic motion of periodic time $T$. Frequency is only factor in simple harmonic motion which does not change with time.
Now, $\frac{a T}{x}=\frac{ m s ^{-2} \times s }{ m }= s ^{-1} \Rightarrow$ unit of frequency
Therefore, $\frac{a T}{x}$ will not change with time.
View full question & answer→MCQ 581 Mark
Two pendulums begin to swing simultaneously. The first pendulum makes nine full oscillations when the other makes seven. The ratio of the lengths of the two pendulums is
- ✓
$\frac{49}{81}$
- B
$\frac{64}{81}$
- C
$\frac{8}{9}$
- D
$\frac{7}{9}$
AnswerCorrect option: A. $\frac{49}{81}$
(a) : Time period of a simple pendulum,
$
T \propto \sqrt{1} \Rightarrow T^2 \propto l
$
Now, $t=n T \Rightarrow t^2=n^2 T^2 \Rightarrow T^2=\frac{t^2}{n^2}$
$
\therefore \quad \frac{T_1^2}{T_2^2}=\frac{l_1}{l_2} \Rightarrow \frac{l_1}{l_2}=\frac{n_2^2 t^2}{n_1^2 t^2} \Rightarrow \frac{l_1}{l_2}=\frac{49}{81}
$
View full question & answer→MCQ 591 Mark
A clock pendulum having coefficient of linear expansion $\alpha=9 \times 10^{-7} /{ }^{\circ} C$ has a period of $0.5 s$ at $20^{\circ} C$. If the clock is used in a climate where the temperature is $30^{\circ} C$, how much time does the clock lose in each oscillation? ( $g=$ constant)
- A
$25 \times 10^{-7} s$
- B
$5 \times 10^{-7} s$
- C
$1.125 \times 10^{-6} s$
- ✓
$2.25 \times 10^{-6} s$
AnswerCorrect option: D. $2.25 \times 10^{-6} s$
(d) : Given, coefficient of linear expansion $(\alpha)$ $=9 \times 10^{-7} /{ }^{\circ} C$
Change in temperature $(\Delta \theta)=30^{\circ} C -20^{\circ} C =10^{\circ} C$
If time period of the clock was $0.5 s$ at $20^{\circ} C$, then time lost by the clock in each oscillation is,
$
\begin{gathered}
{\left[T=2 \pi \sqrt{\frac{l}{g}} \text { taking and differentiation on both sides }\right]} \\
\Delta T=\frac{1}{2} \alpha \Delta \theta T \\
\Delta T=\frac{1}{2} \times 9 \times 10^{-7} \times 10 \times 0.5=2.25 \times 10^{-6} s
\end{gathered}
$
View full question & answer→MCQ 601 Mark
The equation of simple harmonic progressive wave is given by $Y=a \sin 2 \pi(b t-c x)$. The maximum particle velocity will be twice the wave velocity if
- A
$c=\pi a$
- B
$c=\frac{1}{2 \pi a}$
- ✓
$c=\frac{1}{\pi a}$
- D
$c=2 \pi a$
AnswerCorrect option: C. $c=\frac{1}{\pi a}$
(c) : Given, $y=a \sin 2 \pi(b t-c x)$. . . . .(i)
Comparing equation (i) with the standard wave equation $y=a \sin (\omega t-k x)$ we get, $\omega=2 \pi b, k=2 \pi c$
Wave velocity $=\frac{\omega}{k}=\frac{b}{c}$. . . . .(ii)
Particle velocity, $v_P=\frac{d y}{d t}=2 \pi a b \cos (2 \pi b t-2 \pi c x)$
Maximum particle velocity, $\left(v_p\right)_{\max }=2 \pi a b$. . . . .(iii)
According to given problem
$
\begin{aligned}
& \left(v_p\right)_{\max }=2\left(\frac{b}{c}\right)=2 \pi a b \text { (Using (ii) and (iii)) } \\
& c=\frac{1}{\pi a}
\end{aligned}
$
View full question & answer→MCQ 611 Mark
A mass is suspended from a vertical spring which is executing. $\text{S.H.M.}$ of frequency $5\ Hz$. The spring is unstretched at the highest point of oscillation. Maximum speed of the mass is $[$acceleration due to gravity $g=10\ m / s ^2 ]$
- A
$2 \pi\ m / s$
- B
$\pi\ m / s$
- C
$\frac{1}{2 \pi}\ m / s$
- ✓
$\frac{1}{\pi}\ m / s$
AnswerCorrect option: D. $\frac{1}{\pi}\ m / s$
$T=2 \pi \sqrt{\frac{m}{k}}$
$v=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} ;$
$25=\frac{1}{4 \pi^2} \frac{k}{m}$
$\therefore k=100 \pi^2 m$
At equilibrium,
$k A=m g \quad(x=A)$
$A=\frac{m g}{k}$
$v_{\text {max }}=\omega A=2 \pi v A$
$=\frac{10 \pi \times m \times 10}{100 \pi^2 m}$
$=\frac{1}{\pi} ms ^{-1}$
View full question & answer→MCQ 621 Mark
For a particle performing linear S.H.M., its average speed over one oscillation is ( $a=$ amplitude of S.H.M., $n=$ frequency of oscillation)
- A
$2 a n$
- ✓
$4 a n$
- C
$6 a n$
- D
$8 a n$
AnswerCorrect option: B. $4 a n$
(b) : Distance travelled in one oscillation is $4 a$ and time period is $T$.
$
\text { Velocity }=\frac{4 a}{T}=4 a n \quad\left[\because n=\frac{1}{T}\right]
$
View full question & answer→MCQ 631 Mark
The path length of oscillation of a simple pendulum of length 1 meter is $16 cm$. Its maximum velocity is $\left(g=\pi^2 m / s ^2\right)$
- A
$2 \pi cm / s$
- B
$4 \pi cm / s$
- ✓
$8 \pi cm / s$
- D
$16 \pi cm / s$
AnswerCorrect option: C. $8 \pi cm / s$
(c) : Amplitude, $a=\frac{\text { Path length }}{2}=\frac{16}{2}=8 cm$ Time period, $T=2 \pi \sqrt{\frac{l}{g}}=2 \pi \sqrt{\frac{1}{\pi^2}}=2 s$ Maximum velocity, $v_{\max }=a \omega=a \times \frac{2 \pi}{T}=8 \times \frac{2 \pi}{2}$ $=8 \pi cm / s$
View full question & answer→MCQ 641 Mark
A particle is performing S.H.M. starting from extreme position. Graphical representation shows that, between displacement and acceleration, there is a phase difference of
- A
$0 rad$
- B
$\frac{\pi}{4} rad$
- C
$\frac{\pi}{2} rad$
- ✓
$\pi rad$
AnswerCorrect option: D. $\pi rad$
(d) : Acceleration of a particle performing S.H.M. is $a=-\omega^2 x$ or $a \propto-x$
$\therefore \quad$ There is a phase difference of $\pi$ between displacement and acceleration.
View full question & answer→MCQ 651 Mark
A particle performs linear S.H.M. At a particular instant, velocity of the particle is $u$ and acceleration is $\alpha$ while at another instant velocity is $v$ and acceleration is $\beta(0<\alpha<\beta)$. The distance between the two positions is
- ✓
$\frac{u^2-v^2}{\alpha+\beta}$
- B
$\frac{u^2+v^2}{\alpha+\beta}$
- C
$\frac{u^2-v^2}{\alpha-\beta}$
- D
$\frac{u^2+v^2}{\alpha-\beta}$
AnswerCorrect option: A. $\frac{u^2-v^2}{\alpha+\beta}$
(a) : Let $x_1$ be distance travelled by the particle having velocity $u$ and acceleration $\alpha$. Let $x_2$ be the distance travelled by the particle having velocity $v$ and acceleration $\beta$.
If $\omega$ be the angular frequency then
$
\begin{aligned}
& \alpha=\omega^2 x_1 \\
& \beta=\omega^2 x_2 \\
& \text { or } \quad(\alpha+\beta)=\omega^2\left(x_1+x_2\right) .....(i)\\
& u^2=\omega^2 A^2-\omega^2 x_1^2 \\
& v^2=\omega^2 A^2-\omega^2 x_2^2
\end{aligned}
$
$\begin{aligned} \Rightarrow \quad v^2-u^2 & =\omega^2\left(x_1{ }^2-x_2{ }^2\right) \\ v^2-u^2 & =\omega^2\left(x_1+x_2\right)\left(x_1-x_2\right) \\ v^2-u^2 & =\left(x_1-x_2\right)(\alpha+\beta).....(using (i)) \\ \text { or } x_1-x_2 & =\frac{v^2-u^2}{(\alpha+\beta)} \text { or } x_2-x_1=\frac{u^2-v^2}{\alpha+\beta}\end{aligned}$
View full question & answer→MCQ 661 Mark
A particle performing S.H.M. starts from equilibrium position and its time period is 16 second. After 2 seconds its velocity is $\pi m / s$. Amplitude of oscillation is $\left(\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right)$
- A
$2 \sqrt{2} m$
- B
$4 \sqrt{2} m$
- C
$6 \sqrt{2} m$
- ✓
$8 \sqrt{2} m$
AnswerCorrect option: D. $8 \sqrt{2} m$
(d) : The displacement of the particle performing
S.H.M. is given as
$
x=A \sin \omega t
$
Velocity of the particle $v=\frac{d x}{d t}=A \omega \cos \omega t$
Given, $v=\pi m / s , T=16$ second, $t=2$ second
$
\begin{aligned}
& \omega=\frac{2 \pi}{T}=\frac{2 \pi}{16}=\frac{\pi}{8} rad / s ; \pi=A \times \frac{\pi}{8} \cos \frac{\pi}{8} \cdot 2 \\
& \pi=A \times \frac{\pi}{8} \times \cos \frac{\pi}{4} \text { or } 1=\frac{A}{8} \times \frac{1}{\sqrt{2}} \Rightarrow A=8 \sqrt{2} m
\end{aligned}
$
View full question & answer→MCQ 671 Mark
A simple pendulum of length $L$ has mass $M$ and it oscillates freely with amplitude $A$. At extreme position, its potential energy is $( g=$ acceleration due to gravity$)$
- ✓
$\frac{M g A^2}{2 L}$
- B
$\frac{M g A}{2 L}$
- C
$\frac{M g A^2}{L}$
- D
$\frac{2 M g A^2}{L}$
AnswerCorrect option: A. $\frac{M g A^2}{2 L}$
$\text {(a) : The potential energy }=\frac{1}{2} M \omega^2 A^2$
$=\frac{1}{2} M \frac{g}{l} \cdot A^2$
$=\frac{M g A^2}{2 L} \left(\because \omega=\sqrt{\frac{g}{l}}\right)$
View full question & answer→MCQ 681 Mark
A simple pendulum of length $l$ has maximum angular displacement $\theta$. The maximum kinetic energy of the bob of mass $m$ is ( $g=$ acceleration due to gravity)
AnswerCorrect option: C. $m g l(1-\cos \theta)$
(c) : Refer to the figure.
Maximum kinetic energy of the bob $=$ total energy of the bob $=$ maximum potential energy of the bob
$
\begin{aligned}
& =m g h=m g(l-l \cos \theta) \\
& =m g l(1-\cos \theta)
\end{aligned}
$
View full question & answer→MCQ 691 Mark
Which of the following quantity does not change due to damping of oscillations?
Answer(c) : Due to damping of oscillations, time period increases, angular frequency and amplitude decrease but initial phase does not change.
View full question & answer→MCQ 701 Mark
The bob of a simple pendulum performs S.H.M. with period $T$ in air and with period $T_1$ in water. Relation between $T$ and $T_1$ is (neglect friction due to water, density of the material of the bob $=\frac{9}{8} \times 10^3 kg / m ^3$, density of water $1 g / cc$ )
- ✓
$T_1=3 T$
- B
$T_1=2 T$
- C
$T_1=T$
- D
$T_1=\frac{T}{2}$
AnswerCorrect option: A. $T_1=3 T$
(a) : Let $V$ be the volume of the bob, then its weight in air is $W_a=V \rho_b g$
When immersed in water, buoyancy force, $F_b=V \rho_w g$
$\therefore$ Effective weight in water, $W=W_a-F_b$
$
=V \rho_b g-V \rho_m g=V \rho_b g-V \frac{8}{9} \rho_b g=\frac{1}{9} V \rho_b g
$
$\therefore \quad$ Effective acceleration will be $g^{\prime}=\frac{g}{9}$
Now, $T=2 \pi \sqrt{\frac{l}{g}}$ and $T_1=2 \pi \sqrt{\frac{l}{g^{\prime}}}$
$
\Rightarrow \frac{T}{T_1}=\sqrt{\frac{g^{\prime}}{g}}=\sqrt{\frac{1}{9}}=\frac{1}{3} \text { or } T_1=3 T
$
View full question & answer→MCQ 711 Mark
A mass $m_1$ connected to a horizontal spring performs $\text{S.H.M.}$ with amplitude $A$. While mass $m_1$ is passing through mean position another mass $m_2$ is placed on it so that both the masses move together with amplitude $A_1$. The ratio of $\frac{A_1}{A}$ is $m_{2 <}m_1$
- ✓
$\left[\frac{m_1}{m_1+m_2}\right]^{\frac{1}{2}}$
- B
$\left[\frac{m_1+m_2}{m_1}\right]^{\frac{1}{2}}$
- C
$\left[\frac{m_2}{m_1+m_2}\right]^{\frac{1}{2}}$
- D
$\left[\frac{m_1+m_2}{m_2}\right]^{\frac{1}{2}}$
AnswerCorrect option: A. $\left[\frac{m_1}{m_1+m_2}\right]^{\frac{1}{2}}$
Applying principle of momentum conservation at the mean position,
$p_i=p_f$
$ \Rightarrow m_1 v=\left(m_1+m_2\right) v_1$
$ \Rightarrow \frac{v}{v_1}=\frac{m_1+m_2}{m_1}...(i)$
Also$, \frac{v}{v_1}=\frac{\omega A}{\omega_1 A_1}$
$\therefore \frac{A_1}{A}=\frac{\omega}{\omega_1} \times \frac{v_1}{v}....(ii)$
And $\frac{\omega}{\omega_1}=\sqrt{\frac{k}{m_1}} / \sqrt{\frac{k}{m_1+m_2}}=\sqrt{\frac{m_1+m_2}{m_1}}...(iii)$
From eqns. $(i), (ii)$ and $(iii),$
$\frac{A_1}{A}=\sqrt{\frac{m_1+m_2}{m_1}} \times \frac{m_1}{m_1+m_2}=\sqrt{\frac{m_1}{m_1+m_2}}$
View full question & answer→MCQ 721 Mark
A simple pendulum is oscillating with amplitude ' $A$ ' and angular frequency ' $\omega$ '. At displacement ' $x$ ' from mean position, the ratio of kinetic energy to potential energy is
- A
$\frac{x^2}{A^2-x^2}$
- B
$\frac{x^2-A^2}{x^2}$
- ✓
$\frac{A^2-x^2}{x^2}$
- D
$\frac{A-x}{x}$
AnswerCorrect option: C. $\frac{A^2-x^2}{x^2}$
(c) : For a particle executing SHM,
Kinetic energy at displacement $x$ from mean position,
$
K E=\frac{1}{2} m \omega^2\left(A^2-x^2\right)
$
At the same time, its potential energy
$
\begin{aligned}
& P E=\frac{1}{2} m \omega^2 x^2 \\
& \text { Required ratio }=\frac{K E}{P E}=\frac{\frac{1}{2} m \omega^2\left(A^2-x^2\right)}{\frac{1}{2} m \omega^2 x^2}=\frac{A^2-x^2}{x^2}
\end{aligned}
$
View full question & answer→MCQ 731 Mark
A mass is suspended from a spring having spring constant ' $K$ ' is displaced vertically and released, oscillates with period ' $T$ '. The weight of the mass suspended is ( $g=$ gravitational acceleration)
AnswerCorrect option: B. $\frac{K T^2 g}{4 \pi^2}$
(b) : Time period of spring - block system is given by $T=2 \pi \sqrt{\frac{M}{K}}$
Squaring both sides, we get
$
T^2=4 \pi^2 \frac{M}{K} ; M=\frac{K T^2}{4 \pi^2}
$
Weight of mass suspended, $W=M g$
$
W=\frac{K T^2 g}{4 \pi^2}
$
View full question & answer→MCQ 741 Mark
A particle is executing S.H.M. of periodic time ' $T$ '. The time taken by a particle in moving from mean position to half the maximum displacement is $\left(\sin 30^{\circ}=0.5\right)$
- A
$\frac{T}{2}$
- B
$\frac{T}{4}$
- C
$\frac{T}{8}$
- ✓
$\frac{T}{12}$
AnswerCorrect option: D. $\frac{T}{12}$
(d) : The displacement of a particle executing S.H.M. at any time $t$ is given by
$
\begin{array}{ll}
y=A \sin \omega t=A \sin \left(\frac{2 \pi t}{T}\right) \\
\frac{A}{2}=A \sin \left(\frac{2 \pi t}{T}\right) & \\
\sin \left(\frac{2 \pi t}{T}\right)=\frac{1}{2} \Rightarrow \sin \left(\frac{2 \pi t}{T}\right)=\sin \left(\frac{\pi}{6}\right) \\
\therefore \quad \frac{2 \pi}{T} \times t=\frac{\pi}{6} \text { or } t=\frac{T}{12}
\end{array}
$
View full question & answer→MCQ 751 Mark
A particle performs S.H.M. with amplitude $25 cm$ and period $3 s$. The minimum time required for it to move between two points $12.5 cm$ on either side of the mean position is
- A
$0.6 s$
- ✓
$0.5 s$
- C
$0.4 s$
- D
$0.2 s$
AnswerCorrect option: B. $0.5 s$
(b) : The displacement of a particle performing SHM, at any time $t$ is given by,
$
y=A \sin \omega t=A \sin \left(\frac{2 \pi t}{T}\right)
$
Here, $A=25 cm , T=3 s , y_1=12.5 cm$
So, $12.5=25 \sin \left(\frac{2 \pi}{3}\right) t_1 \Rightarrow \frac{1}{2}=\sin \left(\frac{2 \pi t_1}{3}\right)$
$
\begin{aligned}
& \sin \left(\frac{\pi}{6}\right)=\sin \left(\frac{2 \pi t_1}{3}\right) \\
\therefore & \frac{\pi}{6}=\frac{2 \pi}{3} t_1 \Rightarrow t_1=\frac{1}{4} s =0.25 s
\end{aligned}
$
Similarly, for other side of mean position, $t_2=0.25 s$ Required time, $t=t_1+t_2=2 t_1=2 \times 0.25=0.5 s$
View full question & answer→MCQ 761 Mark
A block resting on the horizontal surface executes S.H.M. in horizontal plane with amplitude $A$. The frequency of oscillation for which the block just starts to slip is $(\mu=$ coefficient of friction, $g=$ gravitational acceleration)
- ✓
$\frac{1}{2 \pi} \sqrt{\frac{\mu g}{A}}$
- B
$\frac{1}{4 \pi} \sqrt{\frac{\mu g}{A}}$
- C
$2 \pi \sqrt{\frac{A}{\mu g}}$
- D
$4 \pi \sqrt{\frac{A}{\mu g}}$
AnswerCorrect option: A. $\frac{1}{2 \pi} \sqrt{\frac{\mu g}{A}}$
(a) : Let $m$ be mass of the block. When the block is about to slip, then Force of friction $=$ Centrifugal force
$
\begin{aligned}
& \mu m g=m \omega^2 A \\
& \omega^2=\frac{\mu g}{A} \text { or } \omega=\sqrt{\frac{\mu g}{A}}
\end{aligned}
$
As $\omega=2 \pi \nu$
$
\therefore \quad v=\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{\mu g}{A}}
$
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