MCQ

MCQ 11 Mark

A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm/s. The frequency of its oscillation is …….

A
3 Hz
B
4 Hz
C
2 Hz
✓
1 Hz

Answer

Correct option: D.

1 Hz

$1 Hz$

Solution:
$
v_{\max }= A \omega=2 \pi A f \Rightarrow f=\frac{v_{\max }}{2 \pi A }=\frac{31.4}{2 \times 3.14 \times 5}=1 Hz
$

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MCQ 21 Mark

A simple harmonic oscillator consist of a particle of mass m and an ideal spring with spring constant k The particle oscillates with a time period T. The spring is cut into two equal parts. If one part oscillates with the same particle, the time period will be …….

A
$T / 2$
✓
$\frac{ T }{\sqrt{2}}$
C
$\sqrt{2} T$
D
$2 T$

Answer

Correct option: B.

$\frac{ T }{\sqrt{2}}$

$\frac{ T }{\sqrt{2}}$
Solution:
$
T =2 \pi \sqrt{\frac{m}{k}}
$

If the spring is cut into two equal parts, the force constant of each part becomes $2 k$. Therefore,
$
T ^{\prime}=2 \pi \sqrt{\frac{m}{2 k}}=\frac{ T }{\sqrt{2}}
$

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MCQ 31 Mark

The period of oscillation of a simple pendulum is T in a stationary lift. If the lift moves upwards with an acceleration of 8g, the period will ……..

A
remain the same
B
decrease by T/2
✓
increase by T/3
D
none of these

Answer

Correct option: C.

increase by T/3

increase by T/3
Solution :
Thus, the new time period is T/3. Hence the correct option is (d).

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MCQ 41 Mark

Two simple harmonic motions of angular frequencies 100 and 1000 rad/s have the same displacement amplitude. The ratio of their maximum accelerations is ……….

✓
$1: 10$
B
$1: 10^2$
C
$1: 10^3$
D
$1: 10^4$

Answer

Correct option: A.

$1: 10$

1 : 10
Solution :
The magnitude of the maximum acceleration is given by
$
a_{\max }=\omega^2 A \Rightarrow \frac{\left(a_{\max }\right)_1}{\left(a_{\max }\right)_2}=\frac{\omega_1^2}{\omega_2^2}=\left(\frac{100}{1000}\right)^2=\frac{1}{10^2}
$

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MCQ 51 Mark

Which of the following functions represent SHM?
I. $y=\sin \omega t-\cos \omega t$
II. $y=\sin ^3 t$
III. $y=5 \cos \left(\frac{3 \pi}{4}-3 \omega t\right)$
IV. $y=1+\omega t+\omega^2 t^2$

✓
I and III
B
I and II
C
only I
D
I, II and III

Answer

Correct option: A.

I and III

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MCQ 61 Mark

Which one of the following equations of motion represents simple harmonic motion?

A
Acceleration $=-k_0 x+k_1 x^2$
B
Acceleration $=-k(x+a)$
✓
Acceleration $=k(x+a)$
D
Acceleration $= kx$

Answer

Correct option: C.

Acceleration $=k(x+a)$

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MCQ 71 Mark

The phase difference between the instantaneous velocity and acceleration of a particle executing simple harmonic motion is ……..

✓
0.5π
B
π
C
0.707π
D
zero

Answer

Correct option: A.

0.5π

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MCQ 81 Mark

A particle executing simple harmonic motion of amplitude 31.4 cm/s. The frequency of its oscillation is ………

A
3 Hz
B
4 Hz
C
2 Hz
✓
1 Hz

Answer

Correct option: D.

1 Hz

1 Hz
Solution :
$f_0=\omega A =2 \pi v A$ or$f=\frac{v}{2 \pi A }=\frac{31.4}{2 \times 3.14 \times 5}=1 Hz$

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MCQ 91 Mark

A particle executing simple harmonic motion has a kinetic energy $K _0 \cos ^2 \omega t$. The maximum values of the potential energy and the total energy are, respectively ....

A
$k_0 / 2$ and $k_0$
B
$k_0$ and $2 k_0$
✓
$k _0$ and $k _0$
D
0 and $2 k _0$

Answer

Correct option: C.

$k _0$ and $k _0$

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MCQ 101 Mark

The function $\sin ^2(\omega t)$ represents

✓
a simple harmonic motion with a period π/ω
B
a simple harmonic motion with a period 2π/ω
C
a periodic, but not simple harmonic motion with a period π/ω
D
a periodic, but not simple harmonic motion with a period 2π/ω

Answer

Correct option: A.

a simple harmonic motion with a period π/ω

a simple harmonic motion with a period π/ω
Solution :
$\sin ^2 \omega t=\frac{1}{2}(1-\cos 2 \omega t)$ This is an SHM with period $\frac{2 \pi}{2 \omega}=\frac{\pi}{\omega}$. Its equilibrium position is at $\frac{1}{2}$ instead of zero.

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MCQ 111 Mark

Which one of the following statements is true for the speed v and the acceleration a of a particle executing simple harmonic motion?

A
When v is maximum, a is maximum
B
Value of a is zero, whatever may be the value of v
C
When v is zero, a is zero
✓
When v is maximum, a is zero

Answer

Correct option: D.

When v is maximum, a is zero

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MCQ 121 Mark

In forced oscillations of a particle, the amplitude is maximum for a frequency ω1 of the force, while the energy is maximum for a frequency ω2 of the force. Then ……

A
$\omega_1<\omega_2$
B
$\omega_1>\omega_2$ when damping is small and $\omega_1>\omega_2$ when damping is large.
C
$\omega_1>\omega_2$
✓
$\omega_1=\omega_2$

Answer

Correct option: D.

$\omega_1=\omega_2$

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MCQ 131 Mark

A metal bob is suspended from a coiled spring. When set into vertical vibrations on the earth. It oscillates up and down with frequency f If the same experiment is carried out in a satellite circling the Earth the frequency of vibration will be …….

✓
f
B
zero
C
infinite
D
depend on the distance of the satellite from the earth.

Answer

Correct option: A.

f
solution :
The frequency of oscillation of a mass spring system depends only on the mass and the spring constant.

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MCQ 141 Mark

Two simple harmonic motions act on a particle. These harmonic motions are $x = A \cos (\omega t +\delta)$; $y=A \cos (\omega t+\alpha)$ When $\delta=\alpha+\frac{\pi}{2}$, the resulting motion is .......

A
A circle and the actual motion is clockwise
B
an ellipse and the actual motion is counter clockwise
C
a ellipse and the actual motion is clockwise
✓
a circle and the actual motion is counter clockwise

Answer

Correct option: D.

a circle and the actual motion is counter clockwise

a circle and the actual motion is counter clockwise
solution :
$\left.x= A \cos \left(\omega t+\alpha+\frac{\pi}{2}\right)=(- A \sin +\alpha\right) $
$ y= A \cos (\omega t+\alpha)$
Squaring and adding the two equations, We get $x^2+y^2=A^2$. This is an equation of a circle,Hence the resultant motion is circular. The motion is counter clockwise.

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MCQ 151 Mark

Masses $m_A$ and $m_B$ hanging from the ends of strings of lengths $I_A$ and $I_B$ are executing. Simple harmonic motions. If their frequencies are related as $f_A=2 f_B$, then ......

A
$l_{ A }=2 l_{ B }$ and $m_{ A }=m_{ B } / 2$
B
$l_{ A }=4 l_{ B }$ regardless of masses.
✓
$l_{ A }=l_{ B } / 4$ regardless of masses.
D
$l_{ A }=2 l_{ B }$ and $m_{ A }=2 m_{ B }$

Answer

Correct option: C.

$l_{ A }=l_{ B } / 4$ regardless of masses.

$I_A=I_B / 4$ regardless of masses.
Solution:
$f=\frac{1}{2 \pi} \sqrt{\frac{g}{l}}$ Thus, the correct choice is $(c)$

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MCQ 161 Mark

The amplitude of a damped oscillation reduces to one third of its original value a0 in 20s. The amplitude of such oscillation after a period of 40s will be ……..

✓
$a_0 / 9$
B
$a_0 / 6$
C
$a_0 / 2$
D
$a_0 / 27$

Answer

Correct option: A.

$a_0 / 9$

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MCQ 171 Mark

A particle of mass $m$ is executing oscillations about the origin on the $x$-axis. Its potential energy is $V(x)=k x^2$. Where $k$ is a positive constant. If the amplitude of oscillation is $a$, then its time period $T$ is

A
Proportional to $\frac{1}{\sqrt{a}}$
✓
independent of $a$
C
proportional to $\sqrt{a}$
D
proportional to $a^{3 / 2}$

Answer

Correct option: B.

independent of $a$

independent of a
Solution:
Since $V(x)=K x^2$, the motion is simple harmonic. In SHM, the time period is independent of the amplitude of oscillation.

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MCQ 181 Mark

The kinetic energy of a particle, executing SHM, is $16 J$ when it is at its mean position. If the amplitude of oscillations is $25 cm$, and the mass of the particle is $5.12 kg$, the time period of its oscillation is …….

✓
$\pi/5 s$
B
$ 2\pi s$
C
$20\pi s$
D
$5\pi s$

Answer

Correct option: A.

$\pi/5 s$

$\pi/5 s$
solution :
$ E =\frac{1}{2} m \omega^2 A ^2 \text { or } \omega=\sqrt{\frac{2 E }{m A ^2}} \text { } $
$ T =\frac{2 \pi}{\omega}=\pi A \sqrt{\frac{2 m}{ E }}$
$=\pi \times 0.25 \times \sqrt{\frac{2 \times 5.12}{16}}$
$=\frac{\pi}{4} \times \sqrt{0.64}=\frac{\pi}{5} s $

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MCQ 191 Mark

If the displacement of a particle executing SHM, is given by y = 0.30 sin (220t + 0.64) in metre, then the frequency and the maximum velocity of the particle are (t is in seconds)

✓
35 Hz, 66 m/s
B
45 Hz, 66 m/s
C
58 Hz, 113 m/s
D
35 Hz, 132 m/s

Answer

Correct option: A.

35 Hz, 66 m/s

35 Hz, 66 m/s
solution :
Frequency $f=\frac{\omega}{2 \pi}=\frac{220}{6.28}=35 Hz$

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MCQ 201 Mark

A body is executing simple harmonic motion with an angular frequency 2 rod/sec. The velocity of the body at 20mm displacement, when the amplitude of motion is 60mm, is

A
90 mm/s
✓
113 mm/s
C
118 mm/s
D
131 mm/s

Answer

Correct option: B.

113 mm/s

$113 mm / s$
Solution:
$v=\omega \sqrt{ A ^2-y^2}=2 \times \sqrt{(60)^2-(20)^2}=30 \sqrt{2}=113 mm / s$

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MCQ 211 Mark

Two simple pendulums of lengths 0.5m and 2.0m respectively are given small linear displacement in one direction at the same time. They will again be in phase when the pendulum of shorter length has completed …….. oscillations.

A
5
B
3
C
1
✓
2

Answer

Correct option: D.

2
Solution:
The time period of the shorter pendulum is half that of the longer pendulum. Therefore, the pendulums will again be in phase ( at the mean position). When the shorter pendulum has completed 2 oscillations.

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MCQ 221 Mark

A mass m is vertically suspended from a spring of negligible mass; the system oscillates with a frequency n. What will be the frequency of the system, if a mass 4m is suspended from the same spring?

✓
$\frac{n}{2}$
B
$2 n$
C
$\frac{n}{4}$
D
$4 n$

Answer

Correct option: A.

$\frac{n}{2}$

$\frac{n}{2}$
Solution:
$n=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} \quad ; \quad n^{\prime}=\frac{1}{2 \pi} \sqrt{\frac{k}{4 m}}=\frac{n}{2}$

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MCQ 231 Mark

A seconds pendulum is placed in a space laboratory orbiting around the Earth at a height 3R from the Earth’s surface where R is the radius of the Earth. The time period of the pendulum will be ……..

A
zero
B
2/3s
C
4 s
✓
infinite

Answer

Correct option: D.

infinite

infinite
Solution:
In a space laboratory $g=0$. Therefore, $T =2 \pi \sqrt{\frac{l}{g}}=\infty$

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MCQ 241 Mark

A particle is oscillating according to the equation x = 5 cos (0.5 π t) where t is in seconds. The particle moves from the position of equilibrium to the position of maximum displacement in time ………

✓
1 s
B
2 s
C
0.5 s
D
4 s

Answer

Correct option: A.

1 s

1 s
solution :
$ T =\frac{2 \pi}{\omega}=\frac{2 \pi}{0.5 \pi}=4 s $ Time taken to move from the position of equilibrium to the position of maximum displacement is $t =\frac{T}{4}=1 s$

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MCQ 251 Mark

The angular velocity and the amplitude of a simple pendulum are ω and a, respectively. The ratio of its kinetic and potential energies at a displacement x from the mean position is ……….

A
$\frac{x^2 \omega^2}{a^2-x^2 \omega^2}$
B
$\frac{x^2}{a^2-x^2}$
C
$\frac{a^2-x^2 \omega^2}{x^2 \omega^2}$
✓
$\frac{a^2-x^2}{x^2}$

Answer

Correct option: D.

$\frac{a^2-x^2}{x^2}$

$\frac{a^2-x^2}{x^2}$
Solution:
Kinetic energy $K =\frac{1}{2} m \omega^2\left(a^2-x^2\right)$
Potential energy $U =\frac{1}{2} m \omega^2 x^2 \Rightarrow \frac{ K }{ U }=\frac{a^2-x^2}{x^2}$

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MCQ 261 Mark

A particle, moving along the x-axis, executes simple harmonic motion when the force acting on it is given by (A and k are positive constants.) …….

✓
– Akx
B
A cos (kx)
C
A exp (- kx)
D
Akx

Answer

Correct option: A.

– Akx

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MCQ 271 Mark

The motion of a particle is expressed by the equation a = -bx, where x is the displacement from the mean position, a is the acceleration and b is a constant. The periodic time is ………

A
$\frac{2 \pi}{b}$
✓
$\frac{2 \pi}{\sqrt{b}}$
C
$2 \pi \sqrt{b}$
D
$2 \sqrt{\frac{\pi}{b}}$

Answer

Correct option: B.

$\frac{2 \pi}{\sqrt{b}}$

$\frac{2 \pi}{\sqrt{b}}$=
Solution:
Here $\omega^2=b$. Therefore $T =\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{b}}$

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MCQ 281 Mark

When a particle oscillates simple harmonically, its potential energy varies periodically. If the frequency of oscillation of the particle is n, the frequency of potential energy variation is ……

A
$\frac{n}{2}$
B
$n$
✓
$2 n$
D
$4 n$

Answer

Correct option: C.

$2 n$

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MCQ 291 Mark

A particle is executing simple harmonic motion given by $x=5 \sin \left(4 t-\frac{\pi}{6}\right)$. The velocity of the particle when its displacement is 3 units is .......

A
$\frac{2 \pi}{3}$ units
B
$\frac{5 \pi}{6}$ units
C
20 units
✓
16 units

Answer

Correct option: D.

16 units

16 units
solution :
$v=\omega \sqrt{ A ^2-y^2}=4 \sqrt{25-9}=4 \times 4=16$ units

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MCQ 301 Mark

For a simple pendulum the graph between the length and time period will be a ……….

A
hyperbola
✓
Parabola
C
Straight line
D
none of these

Answer

Correct option: B.

Parabola

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MCQ 311 Mark

A massless spring, having force constant k, oscillates with a frequency n when a mass m is suspended from it. The spring is cut into two equal halves and a mass 2m is suspended from one of the parts. The frequency of oscillation will now be For a simple pendulum the graph between length and time period will be …….

✓
n
B
$n \sqrt{2}$
C
$\frac{n}{\sqrt{2}}$
D
2n

Answer

Correct option: A.

n
solution :
$n=\frac{1}{2 \pi} \sqrt{\frac{k}{m}} ; \quad n^{\prime}=\frac{1}{2} \sqrt{\frac{2}{2}}=n$

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MCQ 321 Mark

When the potential energy of a particle executing simple harmonic motion is one-fourth of its maximum value during the oscillation, the displacement of the particle from the equilibrium position in terms of its amplitude a is ……..

A
$\frac{a}{4}$
B
$\frac{a}{3}$
✓
$\frac{a}{2}$
D
$\frac{2 a}{3}$

Answer

Correct option: C.

$\frac{a}{2}$

$\frac{a}{2}$
Solution:
$\frac{1}{2} m \omega^2 y^2=\frac{1}{4}\left(\frac{1}{2} m\omega^2 a^2\right) \Rightarrow y=\frac{a}{2}$

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MCQ 331 Mark

The amplitude of a damped oscillator becomes half in one minute. The amplitude after 3 minutes will be $\frac{1}{x}$ time the original, where $x$ is .........

A
2 × 3
✓
$2^3$
C
$3^2$
D
$3 \times 2^2$

Answer

Correct option: B.

$2^3$

$2^3$
Solution:
The amplitude decreases exponentially with time and becomes half in 1 minute. Amplitude after 3 minutes $=\left(\frac{1}{2}\right)^3$ of the original value. Thus, $x=2^3$

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MCQ 341 Mark

The equation of SHM of a particle is $\frac{d^2 y}{d t^2}+k y=0,>$ where $k$ is a positive constant. The time period of motion is given by .......

✓
$\frac{2 \pi}{\sqrt{k}}$
B
$\frac{2 \pi}{k}$
C
$\frac{k}{2 \pi}$
D
$\frac{\sqrt{k}}{2 \pi}$

Answer

Correct option: A.

$\frac{2 \pi}{\sqrt{k}}$

$\frac{2 \pi}{\sqrt{k}}$
Solution:
Here $k$ is same as $\omega^2$

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MCQ 351 Mark

A girl is swinging on a swing in the sitting position. How will the period of swing be affected if she stands up?

✓
The period will now be shorter
B
The period will now be longer
C
The period will remain unchanged
D
The period may become longer or shorter depending upon the height of the girl

Answer

Correct option: A.

The period will now be shorter

The period will now be shorter
Solution:
The effective value of I will decrease. Therefore, the time period will be shorter.

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MCQ 361 Mark

The velocity of a particle, undergoing SHM is v at the mean position. If its amplitude is doubled, the velocity at the mean position will be …….

✓
2 v
B
3 v
C
$2 \sqrt{2} v$
D
4v

Answer

Correct option: A.

2 v

2 v $v= A \omega ; v^{\prime}=(2 A ) \omega=2 v$

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MCQ 371 Mark

The maximum displacement of a particle executing SHM is $1 cm$ and the maximum acceleration is $(1.57)^2 cm / s ^2$. Its time period is .....

A
0.25s
✓
4.0s
C
1.57s
D
3.14s

Answer

Correct option: B.

4.0s

4.0s
Solution:
$T =2 \pi \sqrt{\frac{ A }{a_{\max }}}=2 \times 3.14 \times \sqrt{\frac{1.0}{(1.57)^2}}=4.0 s$

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MCQ 381 Mark

A body executes SHM with an amplitude A. Its energy is half kinetic and half potential when the displacement is ……..

A
$\frac{ A }{3}$
B
$\frac{ A }{2}$
✓
$\frac{ A }{\sqrt{2}}$
D
$\frac{ A }{2 \sqrt{2}}$

Answer

Correct option: C.

$\frac{ A }{\sqrt{2}}$

$
\frac{ A }{\sqrt{2}}
$
Solution:
Potential Energy $=\frac{1}{2}$ (Total energy $)$
$
\frac{1}{2} m \omega^2 y^2=\frac{1}{2}\left(\frac{1}{2} m \omega^2 A ^2\right) \Rightarrow y=\frac{ A }{\sqrt{2}} \text {. }
$

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MCQ 391 Mark

A particle executing SHM has an acceleration of $64 cm / s ^2$ with its displacement is $4 cm$. Its time period, in seconds is ......

✓
$\frac{\pi}{2}$
B
$\frac{\pi}{4}$
C
$\pi$
D
$2 \pi$

Answer

Correct option: A.

$\frac{\pi}{2}$

$\frac{\pi}{2}$
Solution:
$
a=\omega^2 y \Rightarrow \omega=\sqrt{\frac{a}{y}} \Rightarrow T =2 \pi \sqrt{\frac{y}{a}}=2 \pi \sqrt{\frac{4}{64}}=\frac{\pi}{2}$

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MCQ 401 Mark

The vertical extension in a light spring by a weight of 1 kg, in equilibrium is 9.8 cm. The period of oscillation of the spring, in seconds, will be ……

✓
$\frac{2 \pi}{10}$
B
$\frac{2 \pi}{100}$
C
$20 \pi$
D
$200 \pi$

Answer

Correct option: A.

$\frac{2 \pi}{10}$

$\frac{2 \pi}{10}$
Solution:
$T =2 \pi \sqrt{\frac{l}{g}}=2 \pi \sqrt{\frac{9.8 \times 10^{-2}}{9.8}}=\frac{2 \pi}{10} s$

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MCQ 411 Mark

A loaded spring vibrates with a period T. The spring is divided into four equal parts and the same load is suspended from one as these parts. The new time period is ………

A
$\frac{ T }{4}$
✓
$\frac{ T }{2}$
C
$2 T$
D
$4 T$

Answer

Correct option: B.

$\frac{ T }{2}$

$\frac{T}{2}$
Solution:
Let the force constant of the spring be $k$. Then $T =2 \pi \sqrt{\frac{m}{k}}$
If the spring is divided into four equal parts, then the force constant of each part will be $4 k$.
Therefore, $T ^{\prime}=2 \pi \sqrt{\frac{ M }{4 k}}=\frac{ T }{2}$

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MCQ 421 Mark

The displacement equation of an oscillator is y = 5 sin (0.2 7 πt + 0.5 π) in SI units. The time period of oscillation is

✓
10 s
B
1 s
C
0.2 s
D
0.5 s

Answer

Correct option: A.

10 s

10 s
Solution:
Comparing with the standard equation $y=A \sin (\omega t+\varphi)$ We have, $\omega=0.2 \pi \Rightarrow T =\frac{2 \pi}{0.2 \pi}=10 s$

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MCQ 431 Mark

The equation $\frac{d^2 y}{d t^2}+b \frac{d y}{d t}+\omega^2 y=0$ represents the equation of motion for a ....... vibration

A
free
✓
damped
C
forced
D
resonant

Answer

Correct option: B.

damped

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MCQ 441 Mark

The frequency of a vibrating body situated in air …….

A
is the same as its natural frequency
B
is higher than its natural frequency
✓
is lower than its natural frequency
D
can have any value

Answer

Correct option: C.

is lower than its natural frequency

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MCQ 451 Mark

The amplitude of a vibrating body situated in a resisting medium ………

A
decreases linearly with time
✓
decrease exponentially with time
C
decreases with time in some other manner
D
remains constant with time

Answer

Correct option: B.

decrease exponentially with time

decreases exponentially with time

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MCQ 461 Mark

A particle is executing SHM. Then the graph of velocity as a function of displacement is …….

A
straight line
B
circle
✓
ellipse
D
hyperbola

Answer

Correct option: C.

ellipse

ellipse
Solution:
$= A \omega \sqrt{1-\frac{y^2}{ A ^2}} \Rightarrow \frac{y^2}{ A ^2}+\frac{v^2}{( A )^2}=1$

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MCQ 471 Mark

A particle is executing SHM. Then the graph of acceleration as a function of displacement is ……..

✓
straight line
B
circle
C
ellipse
D
hyperbola

Answer

Correct option: A.

straight line

straight line
Solution:
In SHM, F ∝ y ⇒ a ∝ y; Thus the graph is a straight line.

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MCQ 481 Mark

A simple harmonic oscillator has a period of 0.01 s and an amplitude of 0.2 m. The magnitude of the velocity in m/s at the centre of oscillation is ……..

A
20π
✓
40π
C
60π
D
80π

Answer

Correct option: B.

40π

40π
Solution:
Velocity is maximum at the centre of oscillation and is given by $v_0=\frac{2 \pi}{ T } A =\frac{2 \pi \times 0.2}{0.01}=40 \pi m / s$

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MCQ 491 Mark

A simple harmonic oscillator has amplitude A and time period T. Its maximum speed is …….

A
$\frac{4 A }{ T }$
B
$\frac{2 A }{ T }$
C
$\frac{4 \pi A }{ T }$
✓
$\frac{2 \pi A }{ T }$

Answer

Correct option: D.

$\frac{2 \pi A }{ T }$

$\frac{2 \pi A }{ T }$
Solution:
$v_{\max }=\omega A =\frac{2 \pi}{ T } A$

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MCQ 501 Mark

In order to double the period of a simple pendulum ……….

A
its length should doubled
✓
its length should be quadrupled.
C
the mass of its bob should be doubled.
D
the mass of its bob should be quadrupled.

Answer

Correct option: B.

its length should be quadrupled.

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