Oscillations Test

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Oscillations Test - 69

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Weekly Quiz Competition

Question 1
1 / -0

An object of mass m is tied to a string of length l and a veriable horizontal force is appiled on it, which starts at zero and gradully increases (it is pulled extremely slowly so that equilibrium exists al all times) untill the string makes an angle $$\theta$$ with the vertical. Work done by the force F is :

A
$$mgl (1 - \sin { \theta } $$)

B
$$mgl$$

C
$$mgl (1 - \cos { \theta }$$)

D
$$mgl (1 - \tan {\theta }$$)

Solution
Question 2
1 / -0

A particle is performing S.H.M. and at $$t = \dfrac{3T}{4}$$, is at position = $$\dfrac{A}{\sqrt 2}$$ and moving towards the origin. Equilibrium position of the particle is at $$x = 0$$. After $$t = \dfrac{3T}{2}$$ what will be the graph of the particle :-

A

B

C

D
Question 3
1 / -0

Time period of revolution of a satellite very close to earth surface is $$T$$. If an imaginary straight bottomless tunnel is dug through earth and a body is dropped into it, then its time period of oscillation will be

A
$$2T$$

B
$$3T$$

C
$$4T$$

D
$$T$$

Solution
Question 4
1 / -0

A source of sound is moving along a circular orbit of radius $$3\ m$$ with an angular velocity of $$10$$ rads/s. A sound detector located for away from the source is executing SHM along the line BD with amplitude $$BC=CD=6\ m$$. The frequency of oscillation of the detector is $$\cfrac{5}{\pi}$$ per sec. The source is at the point $$A$$ when the detector is at the point $$B$$. If the source emits a continuous sound wave frequency $$340$$ Hz. Find the ratio of maximum to minimum frequencies recorded by the detector. (velocity of sound $$= 330\ m/s$$)

A
$$\cfrac{442}{255}$$

B
$$\cfrac{255}{442}$$

C
$$\cfrac{221}{89}$$

D
$$\cfrac{442}{300}$$

Solution
Question 5
1 / -0

A steamer moves with velocity 3 km/h in and against the direction of river water whose velocity is 2 km/h. Calculate the total time for the total journey if the boat travels 2 km in the direction of a stream and then back to its place:

A
2 hrs

B
2.5 hrs

C
2.4 hrs

D
3 hrs.

Solution

When the streamer go along the flow if speed is $$3km/h+2km/h=5km/h$$
Time taken to travel down stream is $${ t }_{ 1 }=\dfrac { distance }{ effective\quad speed } $$
$$\therefore$$ $${ t }_{ 1 }=\dfrac { 2km }{ 5km/h } =0.4h$$
When the streamer go against the flow it speed is $$3km/h-2km/h=1km/h$$
Time taken to travel upstream is $${ t }_{ 3 }=\dfrac { distance }{ effective\quad speed } $$
$${ t }_{ 2 }=\dfrac { 2km }{ 1km/h } =2h$$
Total time $$=2+0.4=2.4h$$
Question 6
1 / -0

Two SHMs are given by $${ Y }_{ 1 }=Asin(\frac { \pi }{ 2 } t+\phi )$$ and $${ Y }_{ 2 }=Bsin(\frac { 2\pi t }{ 3 } t+\phi )$$.The phase difference between these two after '1' sec is

A
$$\pi $$

B
$$\dfrac { \pi }{ 2 } $$

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

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

Solution

For phase difference

$$\phi_{1} - \phi_{2} = \dfrac {2\pi t}{3} + \phi - \dfrac{\pi t}{2} - \phi$$

=$$ \dfrac{\pi t}{6}$$

SO AT t=1 sec

$$\phi_{1} - \phi_{2} = \dfrac{\pi}{6}$$
Question 7
1 / -0

A point performs simple harmonic oscillation of period $$T$$ and the equation of motion is given by $$x=a\sin{(\omega+\pi/6)}$$. After the elapse of what fraction of the time period the velocity of the point will be equal to half of its maximum velocity ?

A
$$T/3$$

B
$$T/12$$

C
$$T/8$$

D
$$T/6$$

Solution
Question 8
1 / -0

Two particles are executing simple harmonic motion of the same amplitude $$A$$ and frequency $$\omega$$ along the x-axis. Their mean position is separated by distance $${X}_{0}({X}_{0}> A)$$. If the maximum separation between them is $$({X}_{0}+A)$$, the phase difference between their motion is

A
$$\pi /2$$

B
$$\pi /3$$

C
$$\pi /4$$

D
$$\pi /6$$

Solution

$$\begin{array}{l} { x_{ 1 } }=A\sin \left( { \omega t+{ \phi _{ 1 } } } \right) \\ { x_{ 2 } }=A\sin \left( { \omega t+{ \phi _{ 2 } } } \right) \\ { x_{ 1 } }-{ x_{ 2 } }=A\left[ { 2\sin \left[ { \omega t+\dfrac { { { \phi _{ 1 } }+{ \phi _{ 2 } } } }{ 2 } } \right] \sin \left[ { \dfrac { { { \phi _{ 1 } }-{ \phi _{ 2 } } } }{ 2 } } \right] } \right] \\ A=2A\sin \left( { \dfrac { { { \phi _{ 1 } }-{ \phi _{ 2 } } } }{ 2 } } \right) \\ \dfrac { { { \phi _{ 1 } }-{ \phi _{ 2 } } } }{ 2 } =\dfrac { \pi }{ 6 } \\ { \phi _{ 1 } }=\dfrac { \pi }{ 3 } \end{array}$$
Hence,
option $$(B)$$ is correct answer.
Question 9
1 / -0

In the arrangement shown, the solid cylinder of mass m is slightly rolled to the left and released. It starts oscillating on the horizontal surface without slipping. Then time period of oscillation is

A
$$\pi \sqrt { \dfrac { 3K }{ m } } $$

B
$$2\pi \sqrt { \dfrac { 3M }{ 2K } } $$

C
$$2\pi \sqrt { \dfrac { 2K }{ 3M } } $$

D
$$2\pi \sqrt { \dfrac { 3K }{ 2m } } $$

Solution

Torque=$$f(R)$$ $$R\rightarrow$$ Radius of cylinder
$$I\propto = fR$$ moment of inertia for solid cylinder
As no slipping $$I=\dfrac{1}{2}mR^2$$
$$a=R\propto$$
$$\dfrac{1}{2} mR^2\left(\dfrac{a}{R}\right)=fR\longrightarrow \therefore f=\dfrac{1}{2}ma$$
$$ma=-kx-f=-kx-\dfrac{1}{2}ma$$
$$-kx=\dfrac{3}{2}ma\Rightarrow \dfrac{d^2x}{dt^2}+\dfrac{2}{3}\dfrac{k}{m}x=0$$
It is of form $$\dfrac{d^2x}{dt^2}+w^2x=0$$
$$\therefore w=\sqrt{\dfrac{2}{3}\dfrac{k}{m}}$$
$$T=\dfrac{2\pi}{w}=2\pi \sqrt{\dfrac{3}{2}\dfrac{m}{R}}$$
Question 10
1 / -0

A particle moves on the $$X-$$ axis according to the equation $$x=x_0\sin^{2}\omega t$$. The motion is simple harmonic

A
With amplitude $$x_0$$

B
With amplitude $$2\ x_0$$

C
With time period $$\dfrac {2\pi}{\omega}$$

D
With time period $$\dfrac {\pi}{\omega}$$.

Solution