Oscillations Test

Result

Oscillations Test - 45

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The position vector of a particle from origin is given by $$\hat r = A\left( {\hat i\cos \omega t + \hat jsin\omega t} \right)$$. The motion of the particle is

A
simple harmonic

B
on a straight line

C
on a circle

D
with contact acceleration

Solution
Question 2
1 / -0

A particle is subjected to two simple harmonic motions given by:
$$y_1=10 \sin\omega t$$ and $$y_2=5 \sin (\omega t+ \pi)$$ The maximum speed of the particle is:

A
$$\sqrt{10^2+5^2 \omega}$$

B
$$\sqrt{10^2-5^2 \omega}$$

C
$$5\omega$$

D
$$15\omega$$

Solution

$$y = {y_1} + {y_2}$$
$$ \Rightarrow y = 10\sin wt + 5\left( {\sin \omega t + \pi } \right)$$
$$ \Rightarrow y = 5\sin \omega t$$
$$\therefore {V_{\max }} = 5 \omega $$
hence,
option $$(C)$$ is correct answer.
Question 3
1 / -0

The maximum time period of oscillation of a simple pendulum of larger length is :

A
$$Infinity$$

B
$$24\ hours$$

C
$$12\ hours$$

D
$$1\ 1/2\ hours$$

Solution
Question 4
1 / -0

A particle is executing $$SHM$$ along $$x$$-axis given by $$x = A \sin \omega t$$. The average velocity during the time interval $$t = 0$$ to $$t = T$$ is

A
$${4A}/{T}$$

B
$$0$$

C
$${2A}/{T}$$

D
$${A}/{T}$$

Solution
Question 5
1 / -0

A particle executes SHM of time period T. It takes time $$t_1$$ to go from $$x=0$$ to $$x = \frac{a}{2}$$ and $${t_2}$$ from $$x = \frac{a}{2}$$ to $$x=0$$. The ratio of $${t_1}:{t_2}$$ will be:

A
1:1

B
1:2

C
1:3

D
2:1

Solution
Question 6
1 / -0

Two particles executes S.H.M along the same line at the same frequency. They move in opposite direction at the mean position. The phase difference will be :

A
$$2\pi $$

B
$$2\pi /3$$

C
$$\pi $$

D
$$\pi /2$$

Solution

During SHM the position of particle is given by
$$x = A\sin wt$$

Let the position 1st particle is given by
$$x = A_1\sin (wt)$$
and that of 2nd [article
$$x = A_2\sin (wt + \phi)$$
as frequency of both partical is same
the position of mean $$\Rightarrow x=0$$

for 1st particle (A)
$$0 = A\sin wt$$
$$ wt = 0$$ ...(1)
as $$t=0$$ at mean position

for 2nd particle (B)
$$0 = \sin (wt + \phi)$$
$$ wt + \phi = 0$$ or $$\pi$$
as if move in opposite direction
$$wt + \phi = \pi$$

as form (1) $$wt = 0$$
$$\Rightarrow \phi = \pi$$
Hence phase difference $$= \pi$$
Question 7
1 / -0

A uniform straight rod of mass 'm; kg and length L is hinged at one end. It is free to oscillate in vertical plane. A point of mass (m kg) is attached to it at a distance 'x' from the hinge. The value of x for which time period of oscillations will be minimum is (use: $$\sqrt { \frac { 7 }{ 3 } =1.5 } $$)

A
$$2L/3$$

B
$$L/2$$

C
$$L/4$$

D
$$2L/7$$

Solution

For a rigid body oscillating about horizontal axis passing through the point O the time period is given by $$T=2\pi \sqrt{\dfrac{l}{mg(l in)}}$$
cg$$=$$length of centre of gravity.
For real centre$$=l/2$$ from ringed point
For mass of x
Centre of mass of system $$c=\dfrac{mx+ml/2}{2m}$$
$$=\dfrac{(2l+l/2)}{2}$$
Now,
Moment of inertia of system $$=\dfrac{1}{3}ml^2+mx^2$$
$$T=2\pi \sqrt{\dfrac{\dfrac{1}{3}ml^2+mx^2}{(2mg)\left(\dfrac{x+l/2}{2}\right)}}$$
T should be minimum then $$\dfrac{\dfrac{1}{3}ml^2+mx^2}{(2mg)\left(\dfrac{x+l/2}{2}\right)}$$ should be minimum
Let $$=\dfrac{\dfrac{1}{3}l^2+x^2}{\left(\dfrac{x}{3}+\dfrac{l}{4}\right)}$$
Should be minimum
$$x=\dfrac{L}{4}$$.
Question 8
1 / -0

A dog of mass $$m$$ is walking on a pivoted disc of radius $$R$$ and mass $$M$$ in a circle of radius $$R/2$$ with an angular frequency $$n$$. The disc will revolve in opposite direction with frequency :

A
$$\dfrac{mn}{M}$$

B
$$\dfrac{mn}{2M}$$

C
$$\dfrac{2 mn}{M}$$

D
$$\dfrac{2Mn}{M}$$

Solution

Angular momentum of the system should be zero.

Moment of inertia of the dog is $$m{\left( {\dfrac{R}{2}} \right)^2}$$

Angular velocity $$=n$$

$$\therefore$$ Angular momentum $$L = IW = m\dfrac{{{R^2}}}{4}n.......\left( 1 \right)$$

Moment of inertia of disc $$ = \frac{1}{2}M{R^2}$$

Let the disc be rotating with angular velocity $$w$$

$$\therefore$$ Angular momentum $$ = \frac{1}{2}M{R^2}w$$

from $$(1)$$ and $$(2)$$ we get,

$$\frac{1}{2}M{R^2}w = \frac{1}{4}m{R^2}n$$

or $$w = \dfrac{{mn}}{{2M}}$$
Question 9
1 / -0

A string of mass m is fixed at both ends. The fundamental tone oscillations are excited in the strong with angular frequency $$\omega $$ and maximum displacement amplitude A. Find the total energy curtained in the string.

A
$$\frac { 1 }{ 2 } { m\omega }^{ 2 }{ A }^{ 2 }$$

B
$$\frac { 1 }{ 4 } { m\omega }^{ 2 }{ A }^{ 2 }$$

C
$$\frac { 1 }{ 6 } { m\omega }^{ 2 }{ A }^{ 2 }$$

D
$$\frac { 1 }{ 8 } { m\omega }^{ 2 }{ A }^{ 2 }$$

Solution

$$y=A\sin kx\cos \omega t$$ where $$k=\dfrac{2\pi}{\lambda}$$
$$\dfrac{dy}{dt}=-A\omega \sin kx\sin \omega t$$
Kinetic energy of element dx at position x is
$$dk_n=\dfrac{1}{2}\mu dx\left(\dfrac{dy}{dt}\right)^2$$
$$=\dfrac{1}{2}\mu A^2\omega^2\cos^2\omega t\cdot \sin^2kx.dx$$
$$k_n=\displaystyle\int^{x=1}_{x=0}dk_n=\dfrac{1}{4}MA^2\omega^2\cos^2\omega t$$.
Question 10
1 / -0

In the arrangement as shown, the length of rod is L and mass is M and it is connected to smooth pin at O.The rod is placed on smooth horizontal table.The spring constant of both the spring is K.The rod is displaced by a small angle $$\theta$$ to perform small ocillation, then is given by

A
$$\left( \cfrac { 3K }{ M } \right) ^{ 1/2 }$$

B
$$\left( \cfrac { 6K }{ M } \right) ^{ 1/2 }$$

C
$$\left( \cfrac { 4K }{ M } \right) ^{ 1/2 }$$

D
$$\left( \cfrac { 3K }{ 2M } \right) ^{ 1/2 }$$