Oscillations Test

Result

Oscillations Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

The displacement of a particle in SHM is indicted by equation $$y=10sin(20t+\dfrac{\pi} {3})$$ where y is in metres. The value of the maximum velocity of the particle will be?

A
$$100 m/s$$

B
$$150 m/s$$

C
$$200 m/s$$

D
$$400 m/s$$

Solution

Here,

$$Y=A\sin(\omega t+\phi)$$

$$v_{max}=A\omega$$

Comparing,

$$v_{max}=10\times 20=200m/s$$

Option $$\textbf C$$ is the correct answer
Question 2
1 / -0

A body executing SHM has a maximum velocity of $$1\ ms^{-1}$$ and a maximum acceleration of $$4ms^{-1}$$. Its time period of oscillation is

A
$$3.14\ s$$

B
$$1.57\ s$$

C
$$6.28\ s$$

D
$$0.25\ s$$

Solution
Question 3
1 / -0

The time period of oscillation of a particle, that executes SHM, is $$1.2s$$. The time, starting from mean position, at which its velocity will be half of its velocity at mean position is ?

A
$$0.1\ s$$

B
$$0.2\ s$$

C
$$0.4\ s$$

D
$$0.6\ s$$

Solution

Time is zero at extreme position
$$ T = 1.2\,sec $$
$$ u = \omega \sqrt{A^{2}-x^{2}}; u_{max} = \omega A $$
$$ u = \dfrac{u_{max}}{2} = \dfrac{\omega A}{2} = \omega \sqrt{A^{2}-x^{2}} $$
$$ \dfrac{A^{2}}{4} = A^{2}-x^{2}; x = \dfrac{\sqrt{3}}{2}A $$
$$ x = A\,cos\,\omega t $$
$$ \dfrac{\sqrt{3}}{2}A = A\,cos\,,\omega t $$
$$ \omega t = \dfrac{x}{6} $$
$$ t = \dfrac{\pi }{6\times \omega } $$
$$ t = \dfrac{\pi }{6}\times \dfrac{1.2}{2\pi } = 0.1\,sec $$
Question 4
1 / -0

The ratio of velocities of particle in SHM, at displacements $$A/3$$ and $$2A/3$$ is ?

A
$$1:2$$

B
$$2:1$$

C
$$\sqrt{8}:\sqrt{5}$$

D
$$\sqrt{5}:\sqrt{8}$$

Solution

Velocity of a particle executing S.H.M. is given by
$$ v= \omega \sqrt{A62-x^2}$$
$$ \omega = \dfrac{2\pi}{T}$$
Here x = A/3 and 2A/3
when displacement is A/3
$$ v= \dfrac{2\pi}{T} \sqrt{A^2 - (\dfrac{A}{3})^2}$$
$$ v = \dfrac{2\pi}{T}\dfrac{A\sqrt{8}}{3}$$
when displacement is 2A/3
$$ {v}' = \dfrac{2\pi}{T}\sqrt{A^2 -(\dfrac{2A}{3})^2}$$
$$ {v}' = \dfrac{2\pi}{T}\dfrac{A\sqrt{5}}{3}$$
So $$ \dfrac{v}{{v}'} = \sqrt{\dfrac{8}{5}}$$
Question 5
1 / -0

In case of a simple pendulum, time period versus length is depicted by

A

B

C

D

Solution
Question 6
1 / -0

The oscillations represented by curve $$1$$ in the graph are expressed by equation $$x=Asin\omega t$$. The equation for the oscillations represented by curve $$2$$ is expressed as:

A
$$x= 2A sin( \omega t- \pi /2)$$

B
$$x= 2A sin( \omega t + \pi /2)$$

C
$$x= - 2A sin( \omega t- \pi /2)$$

D
$$x= A sin( \omega t- \pi /2)$$

Solution

General equation of S.H.M is x= $$A'\sin (\omega t + \phi)$$......................(1)
Amplitude $$A'=2A$$

For $$\phi$$ in eq. (1)
Putting at $$t=0$$ and $$x=-2A$$ from graph
we get $$\phi =\dfrac{-\pi}{2}$$
So, Required equation is
$$x=2A\sin(\omega -\dfrac{\pi}{2})$$
Question 7
1 / -0

The amplitude of a particle in SHM is 5 cm and its time period is $$\pi$$. At displacement of 3 cm from its mean position, the velocity in centimetres per second will be ?

A
8

B
12

C
6

D
2

Solution
Question 8
1 / -0

The equation of motion of a particle executing SHM is $$(\dfrac{d^2x}{dt^2})+kx=0$$. The time period of the particle will be ?

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

B
$$2\pi$$

C
$$2\pi k$$

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

Solution

The equation of motion,

$$\dfrac{d^2x}{dt^2}+kx=0$$

$$a=-kx$$

$$\omega ^2=k$$ ($$a=-\omega ^2 x$$)

$$\omega =\sqrt{k}$$

$$\dfrac{2\pi}{T}=\sqrt{k}$$

Time period,

$$T=\dfrac{2\pi}{\sqrt{k}}$$

The correct option is A.
Question 9
1 / -0

The time -period of a simple pendulum does not depend on

A
length of the pendulum

B
acceleration due to gravity

C
mass of the bob

D
none of these

Solution
Question 10
1 / -0

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

A
Acceleration =$$-k_0x+k_1x^2$$

B
Acceleration =$$-k(x+a)$$

C
Acceleration =$$k(x+a)$$

D
Acceleration =$$kx$$