Oscillations Test

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Oscillations Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

"The motion of a particle with a restoring force gives oscillatory motion". Which of the following force can be a restoring force for the oscillatory motion.

A
Frictional force

B
A constant force F

C
A time varying force opposite to direction of motion

D
Force due to a spring

Solution

Frictional force is a constant force acting opposite to the direction of motion
A constant force only makes an object to move in one direction
A time varying force makes an object to move in one direction with its speed increasing
Force due to a spring, keeps varying, as the object is moved away from its mean position and thus acts as a restoring force

The correct option is (d)
Question 2
1 / -0

The bye-bye gesture, we do using hands is an example of

A
Periodic motion

B
Oscillatory motion

C
Linear motion

D
Circular motion

Solution

The motion of the hand is an oscillatory motion, as the motion repeats in a finite interval of time and the to and fro motion happens in a given time period

As the hand moves to one extreme, a restoring force acts in the opposite direction to move it back to the mean and then to the other extreme position. Thus, the bye-bye gesture of the hand is an oscillating motion

The correct option is (b)
Question 3
1 / -0

A particle executes simple harmonic motion with a frequency $$'f'$$. The frequency with which its kinetic energy oscillates is?

A
$$\dfrac{f}{2}$$

B
$$f$$

C
$$2f$$

D
$$4f$$

Solution

Let, $$x=A\sin\omega t$$
$$v=\cfrac {dx}{dt}=A\omega \cos \omega t$$
Kinetic energy, $$K=\cfrac {1}{2}mv^2$$
$$\implies K=\cfrac {1}{2}m\omega ^2A^2\cos^2\omega t$$
$$\implies K=\cfrac {1}{2}m\omega^2A^2\left (\cfrac {1+cos2\omega t}{2}\right)$$
$$\implies K=\cfrac {1}{4}m\omega^2A^2(1+\cos^2\omega t)$$
$$\therefore \omega_K=2\omega$$
$$\therefore$$ Frequency of oscillation of K.E $$=2f\quad \left[f=\cfrac {\omega}{2\pi}\right]$$
Question 4
1 / -0

Two masses $$m_1$$ and $$m_2$$ are suspended together by a massless spring of spring constant $$k$$ as shown in the figure. When the masses are in equilibrium, $$m_1$$ is removed without disturbing the system. Find the angular frequency and amplitude of oscillation of $$m_2$$.

A
$$\sqrt{\dfrac{k}{3m_2}}$$.

B
$$\sqrt{\dfrac{k}{m_2}}$$.

C
$$\sqrt{\dfrac{k}{2m_2}}$$.

D
$$\sqrt{\dfrac{k}4{m_2}}$$.

Solution

Here, during oscillation
Mass, $$m = m_2$$
$$\therefore \omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{k}{m_2}}$$
Question 5
1 / -0

Two particle P and Q are executing simple harmonic motion with equal amplitude and frequency. At a distance of half the amplitude one of them is moving away from the mean position while the other is observed at a distance half the amplitude to be moving towards mean position as shown in figure. The phase difference between them is:

A
$$\dfrac{\pi}{2}$$

B
$$\dfrac{\pi}{3}$$

C
$$\dfrac{4\pi}{3}$$

D
$$\dfrac{\pi}{6}$$

Solution

Equation of SHM: $$x=a\sin wt$$
for $$P$$ $$a\sin (wt +\phi)=a/2$$
$$\sin (wt) 1/2$$
let $$t=0$$
$$a\sin \phi=a/2$$
$$\phi=\pi/6$$
for $$Q:a\sin (wt+\delta )=-a/2$$
$$\sin (\delta )=-1/2$$
$$\delta =-\pi /6$$
Phase difference $$=\phi-\delta$$
$$=\dfrac{\pi}{6}-\left(-\dfrac{\pi}{6}\right)$$
$$\boxed{\pi/3}$$
Question 6
1 / -0

Figure shows three systems in which a block of mass m can execute S.H.M. What is ratio of frequency of oscillation?

A
2: 1: 4

B
1: 2: 4

C
4: 2: 1

D
3: 2: 1

Solution

(i) $$\omega_1=\sqrt{\dfrac{K}{m}}$$
(ii) $$k\dfrac{X}{2}=2T$$
$$T= ma = -\dfrac{kx}{4}$$
$$\omega_2 =\sqrt{\dfrac{k}{4m}}$$
$$-4kx = ma$$
$$a=-\dfrac{4k}{m}x, \omega_3$$$$=\sqrt{\dfrac{4k}{m}}$$
Question 7
1 / -0

Two blocks A and B of masses m and 2 m, respectively, are held at rest such that the spring is in natural length. Find out the accelerations of both the blocks just after release.

A
$$g\downarrow ,g\downarrow $$

B
$$\dfrac { g }{ 3 } \downarrow ,\dfrac { g }{ 3 } \uparrow $$

C
$$0, 0$$

D
$$g\downarrow ,c$$

Solution

As spring is in natural length then
$$a = \left(\dfrac{m_1 + m_1}{m_2 + m} \right) g$$
Case (1)
take, $$m_2 = 2m$$
$$m_1 = m$$ then
$$a = \left(\dfrac{2m - m}{3m} \right) g$$
$$a = g/3 \uparrow$$
Case (2)
take $$m_2 = m $$
$$m_1 = 2m$$ then
$$a = \left(\dfrac{m - 2m}{3m} \right) g$$
$$a = -9/3$$ or $$a = 9/3 \downarrow$$
Question 8
1 / -0

A particle executes simple harmonic motion between $$x=-A$$ and $$x=+A$$. The time taken for it to go from $$0$$ to $$A/2$$ is $$T_1$$ and to go from $$A/2$$ to A is $$T_2$$. Then.

A
$$T_1 < T_2$$

B
$$T_1 > T_2$$

C
$$T_1 = T_2$$

D
$$T_1 =2T_2$$

Solution

Using $$x=A \sin \omega t$$
For $$x=A/2,\ sin \omega T_1=1/2 \Rightarrow T_1=\dfrac{\pi}{6 \omega}$$
For $$x=A,\ \sin \omega (T_1+T_2)=1 \Rightarrow T_1+T_2=\dfrac{\pi}{2 \omega}$$
$$\Rightarrow T_2=\dfrac{\pi}{2 \omega}-T_1=\dfrac{\pi}{2 \omega} - \dfrac{\pi}{6 \omega}=\dfrac{\pi}{3 \omega} \ i.e., T_1 < T_2$$
Alternative method : In SHM, velocity of particle also oscillates simple harmonically. Speed is more near the mean position and less near the extreme position. Therefore the time taken for the particle to go from $$0$$ to $$\dfrac{A}{2}$$ will be less than the time taken to go from $$\dfrac{A}{2}$$ to $$A$$. Hence $$T_1 < T_2$$
Question 9
1 / -0

Two bodies M and N of equal mass are suspended from two separate massless spring of force constant $$k_1$$ and $$k_2$$ respectively. If the two bodies oscillate vertically such that their maximum velocities are equal, the ratio of the amplitude of vibration of M to that of N is?

A
$$\dfrac{k_1}{k_2}$$

B
$$\sqrt{\dfrac{k_1}{k_2}}$$

C
$$\dfrac{k_2}{k_1}$$

D
$$\sqrt{\dfrac{k_2}{k_1}}$$

Solution

$$V_{max}=A\omega$$
$$\implies {V_{max}}_1={V_{max}}_{2}$$
$$\implies A_1\omega _1=A_2\omega_2$$
$$\implies \cfrac {A_1}{A_2}=\cfrac {\omega_2}{\omega_1}$$
Now, $$K_1=m\omega_1^2$$

$$\implies \omega_1=\sqrt {\cfrac {K_1}{m}}$$

and $$K_2=m\omega_2^2$$
$$\implies \omega_2=\sqrt {\cfrac {K_2}{m}}$$

$$\therefore \cfrac {A_1}{A_2}=\sqrt {\cfrac {K_2}{K_1}}$$
Question 10
1 / -0

The displacement of a particle performing linear S.H.M is given by $$x=6 sin(3 \pi t-5\pi/6) $$ metre. Find the time at which the particle reaches the extreme position towards the left:

A
5/9 secs

B
4/9 secs

C
1/9 secs

D
7/9 secs

Solution

The amplitude at the extremes are +A and -A respectively. Comparing $$x=6 sin(3 \pi t+5\pi/6) $$ with $$ x=A sin (\omega t + \phi)$$, we get amplitude A = 6.

For the particle to reach the amplitude, we have $$(3 \pi t-5 \pi/6)= \pi/2 \implies 3 \pi t =5 \pi/6 + 3 \pi/6=4 \pi/3$$

Thus, t = 4/9 secs

The correct option is (b)