Oscillations Test

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Oscillations Test - 60

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Weekly Quiz Competition

Question 1
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A particle of mass 200 g executes linear simple harmonic motion with an amplitude 10 cm. When the particles at a point midway between the mean and the extreme position, its kinetic energy is $$3\pi ^{2}\times 10^{-3}J$$. Assuming the initial phase to be $$\dfrac{2\pi }{3}$$, the equation of motion of the particle will be :

A
y=10 sin $$(2\pi t +\dfrac{2\pi }{3})$$cm

B
y=10 sin $$(4\pi t +\dfrac{2\pi }{3})$$cm

C
y=10 cos $$(2\pi t +\dfrac{\pi }{6})$$cm

D
y=10 cos $$(2\pi t +\dfrac{\pi }{3})$$cm

Solution
Question 2
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The angular frequency of a spring block system is $$\omega _{0}$$ This system is suspended from the ceiling of anelevator moving downwards with a constant speed v$$_{0}$$. The block is at rest relative to the elevator. Lift issuddenly stopped. Assuming the downwards as a positive direction, choose the wrong statement:

A
The amplitude of the block is $$\dfrac{v_{0}}{\omega_{0}}$$

B
The initial phase of the block is $$\pi $$

C
The equation of motion for the block is $$\dfrac{v_{0}}{\omega_{0}} \sin \omega _{0}$$t.

D
The maximum speed of the block is v$$_{0}$$.

Solution

Since $${ V }_{ 0 }$$ is the maximum speed of the ball in its equilibrium position.
$$\therefore { V }_{ 0 }={ \omega }_{ 0 }A\\ A=\dfrac { { V }_{ 0 } }{ { \omega }_{ 0 } }$$
So equation of motion becomes $$\dfrac { { V }_{ 0 } }{ { \omega }_{ 0 } } =Asin{ \omega }_{ 0 }t$$ as there was no initial phase in the SHM.
There was no initial phase in the SHM.
Hence B is correct.
Question 3
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A particle of mass 10 gm is placed in a potential field given by $$V = (50x^2 + 100) \ J/kg$$. The frequency of oscillation in cycles/sec is :

A
$$\dfrac{10}{\pi}$$

B
$$\dfrac{5}{\pi}$$

C
$$\dfrac{100}{\pi}$$

D
$$\dfrac{50}{\pi}$$

Solution

Potential energy $$U = mV$$
$$\Rightarrow U = (50x^2+100)10^{-2}$$
$$F = -\dfrac{dU}{dx} = -(100x) 10^{-2}$$
$$\Rightarrow m{\omega}^2x = -(100\times 10^{-2})x$$
$$10\times 10^{-3}{\omega}^2x = 100\times 10^{-2}x$$
$$\Rightarrow {\omega}^2 = 100, {\omega} = 10$$
$$\Rightarrow f = \dfrac{\omega}{2\pi}=\dfrac{10}{2\pi}=\dfrac{5}{\pi}$$
Question 4
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System shown in figure is in equilibrium. Find the magnitude of net change in the string tension between two masses just after, when one of the springs is cut. Mass of both the blocks is same and equal to $$m$$ and spring constant of both the springs is $$k$$

A
$$\displaystyle \frac{mg}{2}$$

B
$$\displaystyle \frac{mg}{4}$$

C
$$\displaystyle \frac{mg}{3}$$

D
$$\displaystyle \frac{3mg}{2}$$

Solution

$$T_i = mg$$
$$2kx = 2mg$$
$$\therefore kx = mg$$
One $$kx$$ force (acting in upward direction) is suddenly removed. So net downward force on system will be $$kx$$ or $$mg$$. Therefore net downward acceleration of system,
$$a = \displaystyle\frac{mg}{2m} = \displaystyle\frac{g}{2}$$
$$\mbox{Free body diagram of lower block gives the equation,}$$
$$mg - T_f = ma = \displaystyle\frac{mg}{2}$$
$$\therefore \quad T_f = \displaystyle\frac{mg}{2}$$
$$\mbox{From these two equations we get,}$$
$$\triangle T = \displaystyle\frac{mg}{2}$$
Question 5
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A block of mass $$m$$ is attached to the spring $$k$$ in free length and released at time $$t=0$$ in the position $$O$$ from rest. For the subsequent vertical motion of the block the equation of motion is given by $$\left ( \omega =\sqrt{\displaystyle \frac{k}{m}} \right )$$

A
$$x=\displaystyle \frac{mg}{k}\sin \omega t$$

B
$$x=\displaystyle \frac{mg}{k}\cos \omega t$$

C
$$x=\displaystyle \frac{mg}{k}(1+ \cos \omega t)$$

D
$$x=\displaystyle \frac{mg}{k}(1- \cos \omega t)$$

Solution

The angular velocity of the system is $$\omega=\sqrt{k/m}$$
Here, the point $$O$$ is the extreme point of SHM.
The amplitude, $$A$$ is given by, $$kA=mg\Rightarrow A=mg/k$$
So the equation of motion from point $$O$$ is $$A-Asin(\omega t+\pi/2)=\dfrac{mg}{k}(1-cos\omega t)$$
Question 6
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The system shown in the given figure is in equilibrium. The magnitude of net change in the string tension between two masses after, when one of the springs is cut is
(Mass of both the blocks is same and equal to $$m$$ and spring constant of both the springs is $$k$$ and acceleration due to gravity $$=g$$)

A
$$\dfrac {mg}{2}$$

B
$$\dfrac {mg}{4}$$

C
$$\dfrac {mg}{3}$$

D
$$\dfrac {3mg}{2}$$

Solution

$$T_i=mg$$
$$2kx=2mg$$
$$\therefore kx=mg$$
One $$kx$$ force (acting in upward direction) is suddenly removed. So net downward force on system will be $$kx$$ or $$mg$$. Therefore net downward acceleration of system,
$$a=\dfrac {mg}{2m}=\dfrac {g}{2}$$
Free body diagram of lower block gives the equation,
$$mg-T_f=ma=\dfrac {mg}{2}$$
$$\therefore T_f=\dfrac {mg}{2} .....(ii)$$
From these two equations we get,
$$\Delta T=\dfrac {mg}{2}$$
Question 7
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Directions For Questions

The equation of a particle moving in SHM is $$\displaystyle \frac{d^2X}{dt^2}=-\omega^2x$$ where $$\omega$$ is a constant, being equal to $$\displaystyle \frac{2\pi}{Time\ period}$$. The velocity of such a particle is maximum when it passes through its mean position while it is subjected to maximum acceleration at the extreme positions. The solution to the above equation is $$x=A\sin{(\omega t+\theta)}$$, where $$\theta$$ is a constant called the initial phase of the motion.

...view full instructions

A particle in SHM has an amplitude of 20 cm and time period of 2 sec. Its maximum velocity will be(in m/s):

A
$$0.04 \ \pi^2$$

B
$$0.2 \ \pi$$

C
$$0.2 \ \pi^2$$

D
$$0.04 \ \pi$$

Solution

$$v=\dfrac{dx}{dt}=A\omega cos(\omega t + \theta)$$
Maximum $$v$$ occurs for maximum value of $$cos(\omega t + \theta)$$, which is 1
Thus, maximum $$v=A\omega$$
Now, $$\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{2}=\pi$$, and $$A = 20 cm = 0.2 m$$
Thus, maximum v = $$0.2\pi$$
Question 8
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A particle is an linear simple harmonic motion between two extreme point A and B. 10 cm apart(See figure below) If the direction from A to B is taken as positive direction, what are signs of displacement x, velocity V and acceleration a, when the particle is at A?

A
$$x=-ve, V=-ve, a=-ve$$

B
$$x=+ve, V=0, a=-ve$$

C
$$x=+ve, V=-ve, a=+ve$$

D
$$x=-ve, V=0, a=+ve$$

Solution

When particle is at extreme A, velocity is zero. Since it is on the negative side of the mean, thus x is negative. Acceleration will be towards right that pulls it back (= - kx), thus acceleration is positive (as x is negative).
Question 9
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The time period of the oscillating system (image above) is

A
$$T = 2\pi\sqrt{\displaystyle\frac{m}{k_1k_2}}$$

B
$$T = 2\pi\sqrt{\displaystyle\frac{m}{k_1+k_2}}$$

C
$$T = 2\pi\sqrt{mk_1k_2}$$

D
None of these

Solution

Let block is displaced by $$X$$ to the right.
Restoring Force by spring 1, $$F_1=k_1X$$
Restoring Force by spring 2, $$F_2=k_2X$$
Net Restoring force$$=(k_1+k_2)X$$
$$\Rightarrow m\ddot{X}= - (k_1+k_2)X$$
$$\Rightarrow \ddot{X}= -\dfrac{(k_1+ k_2)}{m} X$$ $$\Rightarrow \ddot{X}= -\omega^2 X$$
$$\Rightarrow \omega= \sqrt{\dfrac{(k_1+k_2)}{m}}$$
$$\therefore \text{Time Period } T= \dfrac{2\pi}{\omega}$$ $$\Rightarrow T= 2\pi \sqrt{\dfrac{m}{k_1+k_2}}$$
Question 10
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Five identical springs are used in the following three configurations. The time periods of vertical oscillations in configurations (i), (ii) and (iii) are in the ratio

A
$$1:\sqrt{2}:1/\sqrt{2}$$

B
$$2:\sqrt{2}:1/\sqrt{2}$$

C
$$1/\sqrt{2}:2:1$$

D
$$2:1/\sqrt{2}:1$$

Solution

By above formulae, we get,
$$T_{1}=2\pi \sqrt{\frac{m}{k}}$$
$$T_{2}=2\pi \sqrt{\frac{2m}{k}}$$..(Series combination of springs)
$$T_{3}=2\pi \sqrt{\frac{m}{2k}}$$..(Parallel combination of springs)
Taking ratio, we get Answer as option A.