[ 2 Marks Questions ] - Physics STD 11 Questions - Vidyadip

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Question 12 Marks

A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2s. Find the acceleration and velocity of the body when the displacement is 5 cm.

Answer

Here $A=5 cm , T=0.2 s$
Velocity and acceleration at any displacement $x$ are given by
$
\begin{aligned}
v & =\omega \sqrt{ A ^2-x^2}=\frac{2 \pi}{ T } \sqrt{ A ^2-x^2} \\
a & =-\omega^2 x=-\frac{4 \pi^2}{ T ^2} x \\
a & =\frac{2 \pi}{0.2} \sqrt{5^2-5^2}=0 \\
a & =-\frac{4 \pi^2}{(0.2)^2} \times 5 cms ^{-2}=-500 \pi^2 cm s ^{-2}=-5 \pi^2 ms ^{-2}
\end{aligned}
$

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Question 22 Marks

A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes 5T/3. What is the ratio m/M?

Answer

With mass $M$, the time period of the spring is $T =2 \pi \sqrt{\frac{ M }{k}}$ With mass $M +m$, the time period becomes
$
\begin{aligned}
\frac{5 T }{3} & =2 \pi \sqrt{\frac{ M +m}{k}} \text { or } \frac{5}{3} \times 2 \pi \sqrt{\frac{ M }{k}}=2 \pi \sqrt{\frac{ M +m}{k}} \\
\frac{25}{9} M & = M +m \quad ; \quad \frac{16}{9} M =m \quad \text { or } \quad \frac{m}{ M }=\frac{16}{9}
\end{aligned}
$

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Question 32 Marks

A 0.2 kg of mass hangs at the end of a spring. When 0.02 kg more mass is added to the end of the spring, it stretches 7 cm more. If the 0.02 kg mass is removed, what will be the period of vibration of the system?

Answer

When $0.02 kg$ mass is added, the spring streches by $7 cm$
As
$
\begin{aligned}
m g & =k x \\
k & =\frac{m g}{x}=\frac{0.02 \times 10}{7 \times 10^{-2}}=\frac{20}{7} Nm ^{-1}
\end{aligned}
$
When $0.02 kg$ mass ia removed, the period of vibration will be
$
T =2 x \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{0.2}{20 / 7}}=2 \pi \sqrt{\frac{7}{100}}=\frac{2 \pi \times 2.645}{10}=1.66 s
$

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Question 42 Marks

A $5 kg$ collar is attached to a spring of force constant $500 Nm ^{-1}$. It slides without friction on a horizontal rod as shown in figure. The collar is displaced from its equilibrium position by $10.0 cm$ and released.
Calculate:
(i) the period of oscillation
(ii) the maximum speed, and
(iii) the maximum acceleration of the collar.

Answer

Here $m=5 kg , k=500 Nm ^{-1}, A =10.0 cm =0.10 m$
(i) Period of oscillation,
$
T =2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{5}{500}}=2 \times 3.14 \times \frac{1}{10} s =0.628 s
$
(ii) The maximum speed of the collar,
$
v_{\max }=\omega^2 A =\frac{k}{m} A =\frac{500}{5} \times 0.10=10 ms ^{-2}
$
(iii) The maximum acceleration of the collar,
$
a_{\max }=\omega^2 A =\frac{k}{m} A =\frac{500}{5} \times 0.10=10 ms ^{-2}
$

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Question 52 Marks

A block whose mass is $1 kg$ is fastened to a spring. The spring has a spring constant of $50 Nm ^{-1}$. The block is pulled to a distance $x=10 cm$ from its equilibrium position at $t=0$ on a frictionless surface from rest at $t=0$. Calculate the kinetic, potential and total energies of the block when it is $5 cm$ away from the mean position.

Answer

Here $m=1 kg , k=50 Nm ^{-1}, A =10 cm =0.10 m , y=5 cm =0.05 m$
Kinetic energy: $\quad E _{ K }=\frac{1}{2} k\left( A ^2-y^2\right)=\frac{1}{2} \times 50\left[(0.10)^2-(0.05)^2\right]=0.1875 J$
Potential energy, $E _{ P }=\frac{1}{2} k y^2=\frac{1}{2} \times 50 \times(0.05)^2=0.0625 J$
Total energy, $\quad E=E_K+E_p=0.1875+0.0625=0.25 J$

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Question 62 Marks

Two simple harmonic motions are represented by the equations: $x_1=5 \sin \left(2 \pi t+\frac{\pi}{4}\right) ; x^2=5 \sqrt{2}(\sin 2 \pi t+\cos 2 \pi t)$What is the ratio of their amplitudes?

Answer

$ x_1=5 \sin \left(2 \pi t+\frac{\pi}{4}\right)$
${\left[\therefore A_1=5\right]}$
$x_2=5 \sqrt{2}(\sin 2 \pi t+\cos 2 \pi t)$
$=10 \sin \left(\sin 2 \pi t \cos \frac{\pi}{4}+\cos 2 \pi t \sin \frac{\pi}{4}\right)$
$x_2=10 \sin \left(2 \pi t+\frac{\pi}{4}\right)$
${\left[\therefore A _2=10\right]}$
$\frac{ A _1}{ A _2}=\frac{5}{10}=1: 2$
$A_1: A_2=1: 2$
$ $
Hence,

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Question 72 Marks

A body of mass m is attached to lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5s. Find the value of m is kg.

Answer

$
\begin{aligned}
T _1 & =3=2 \pi \sqrt{\frac{m}{k}} \\
T _2 & =5=2 \pi \sqrt{\frac{m+1}{k}} \\
\frac{ T _1}{ T _2} & =\frac{3}{5}=\sqrt{\frac{m}{m+1}} \\
\frac{9}{25} & =\frac{m}{m+1} \Rightarrow 9 m+9=25 m \\
m & =\frac{9}{16} kg
\end{aligned}
$

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Question 82 Marks

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then find its time period in seconds.

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Question 92 Marks

The acceleration due to gravity on the surface of the moon is $1.7 ms ^{-2}$. What is the time period of simple pendulum on the moon if its time period on the earth is $3.5 s$ ? Give $g$ on Earth $=9.8$ $ms ^{-2}$

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Question 102 Marks

A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum oscillator. The acceleration of the bob of the pendulum is $20 ms ^{-2}$ at a distance of $5 m$ from the mean position. To find the time period of oscillation.

Answer

Given; $a=20 ms ^{-2} ; y =5 ma =\omega^2 y$

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Question 112 Marks

Define Amplitude.

Answer

The maximum displacement of the oscillating particle on either side of its mean position is called its amplitude.

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Question 122 Marks

What is meant by angular oscillation?

Answer

When a body is allowed to rotate freely about a given axis then the oscillation is known as the angular oscillation.

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Question 132 Marks

What is meant by phase in SHM?

Answer

Phase: The phase of a vibrating particle at any instant completely specifies the state of the particle. It expresses the position and direction of motion of the particle at that instant with respect to its mean position.

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Question 142 Marks

Define acceleration in SHM.

Answer

Acceleration: The rate of change of velocity is acceleration.
$
\begin{aligned}
a & =\frac{d v}{d t}=\frac{d}{d t}( A \omega \cos \omega t) \\
& a=-\omega^2 A \sin \omega t=-\omega^2 y \\
\therefore \quad a & =\frac{d^2 y}{d t^2}=-\omega^2 y
\end{aligned}
$

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Question 152 Marks

Define velocity in SHM.

Answer

Velocity: The rate of change of displacement is velocity.
$v=\frac{d y}{d t}=\frac{d}{d t}( A \sin \omega t) $
$v=\frac{d y}{d t}= A \omega \cos \omega t$

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Question 162 Marks

What is meant by displacement in SHM?

Answer

The distance travelled by the vibrating particle at any instant of time t from its mean position is known as displacement. y = A sin ωt
The maximum displacement from the mean position is known as amplitude (A) of the vibrating particle.

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Question 172 Marks

What is meant by simple harmonic motion (SHM)?

Answer

A particle is said to execute simple harmonic motion if it moves to and fro about a mean position under the action of a restoring force which is directly proportional to its displacement from the mean position and is always directed towards the mean position.

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Question 182 Marks

What is meant by Oscillatory motion?

Answer

When an object or a particle moves back and forth repeatedly for some duration of time, its – motion is said to be oscillatory (or vibratory).

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[ 2 Marks Questions ] - Physics STD 11 Questions - Vidyadip