Oscillations Test

Result

Oscillations Test - 17

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

A simple harmonic oscillator is of mass $$0.100$$ kg. It is oscillating with a frequency of $$\dfrac{5}{\pi }$$Hz. If its amplitude of vibration is $$5$$ cm, the force acting on the particle at its extreme position is

A
2 N

B
1.5 N

C
1 N

D
0.5 N

Solution

$$ F = m {\omega}^{2} A$$
$$\omega = 2\pi f = 10 rad / sec$$
$$F = 0.1 \times 100\times \dfrac{5}{100}$$
$$ = 0.5 N $$
Question 2
1 / -0

Two particles executing SHM of the same amplitude and frequency on parallel lines side by side. They cross one another when moving in opposite directions each time their displacement is $$\dfrac{\sqrt{3}}{2}$$ times their amplitude. The phase difference between them is:

A
$$30^{\circ}$$

B
$$45^{\circ}$$

C
$$60^{\circ}$$

D
$$90^{\circ}$$

Solution

since $$x=A\sin { \left( \omega t+\phi \right) }$$
$$\\ \dfrac { \sqrt{3}A

}{ 2 } =A\sin { \left( \omega t+\phi \right) } \\ \Rightarrow \sin {

\left( \omega t+\phi \right) } =\dfrac{\sqrt{3}}{2} \\ \Rightarrow \omega

t+\phi =60^{ \circ }or\quad 120^{ \circ }$$
so the one particle has phase of $$60^{ \circ }$$ and another has $$120^{ \circ }$$
So the phase difference is $$120^{ \circ }-60^{ \circ }=60^{ \circ }$$
Question 3
1 / -0

A particle is in S.H.M of amplitude $$ 2$$ cm. At extreme position the force is $$4$$N. At the point mid-way between mean and extreme position, the force is :

A
$$1$$ N

B
$$2$$N

C
$$3$$N

D
$$4$$N

Solution

Amplitude = 2 cm
Force = 4N
$$F = mw^{2}A=4$$
$$F_{1} = mw^{2}x$$
Since $$F$$ is directly proportional to $$x$$ so , at midpoint the force when the amplitude is $$2 \ cm$$ will be $$2N$$
Question 4
1 / -0

The average speed of S.H.M oscillator averaged over $$\dfrac{3T}{8}$$ is

A
$$\dfrac{4\sqrt{2}}{3} \dfrac{A\omega }{\pi }$$

B
$$\dfrac{2A(4-\sqrt{2})}{3\pi }$$

C
$$\dfrac{3A\omega 4- \sqrt{2}\omega }{4\pi }$$

D
$$\dfrac{5\sqrt{2\pi }}{2\omega }$$
Question 5
1 / -0

The net external force acting on the disc when its centre of mass is at displacement $$x$$ with respect to its equilibrium position is:

A
$$-kx$$

B
$$-2kx$$

C
$$-\dfrac{2kx}{3}$$

D
$$\dfrac{4kx}{3}$$

Solution

$$\textbf{Given:- }$$ The net external force acting on the disc

$$\textbf{Solution:-}$$ Net Force acting on a body is given by the formula:
$$\vec{F_{net}} = m.\vec{a}$$
Where,
$$\vec{F_{net}}$$ is the net force acting on the body

m is the mass of the body

$$\vec{a}$$ is the acceleration of the body

We will insert$$\left ( -2kx + F \right )$$ in the place of $$\vec{F_{net}}$$

$$-2kx + F = Ma_{c}................(1)$$

Where,
k is the spring constant
x is the distance by which the spring has been stretched
F is the friction force on the disk
M is the mass of disk
$$a_{c}$$ is the acceleration of centre of mass of the disk

Net Torque acting on a body is given by the formula,

$$\vec{\tau _{net}} = I\vec{\alpha }...............(2)$$

Where,

$$\vec{\tau _{net}}$$ is the net Torque acting on the body

I is the moment of inertia of the body

$$\vec{\alpha }$$ is the angular acceleration of the body

Additionally, Torque is also calculated as follows,
$$\vec{\tau _{net}} = \vec{F}\times R................(3)$$

Where,
R is the distance of the Force from the center of mass of the body

Now combining equations 2 and 3,

$$\vec{F}\times R = T\vec{\alpha}$$

$$\vec{F} = \dfrac{T\vec{\alpha}}{R}.................(4)$$

In Pure Rolling condition,

$$a_{c} = \vec{\alpha_{c}}R$$

Moment of inertia of a disk is given by the formula,

F $$= \dfrac{\dfrac{1}{2}MR^{2}}{R}\times \dfrac{a_{c}}{R}$$

$$\Rightarrow F = \dfrac{1}{2}Ma_{c}$$

Inserting the value of F in equation 1,

$$\Rightarrow -2kx + \dfrac{1}{2}Ma_{c} = -Ma_{c}$$

$$\Rightarrow \dfrac{1}{2}Ma_{c} + Ma_{c} = 2kx $$

$$\Rightarrow \dfrac{3}{2}Ma_{c} = 2kx$$

$$\Rightarrow Ma_{c} = \dfrac{4}{3}kx$$

$$\Rightarrow MA_{c} = \dfrac{4}{3}kx$$

$$\textbf{Hence the correct option is D}$$
Question 6
1 / -0

A particle executes SHM on a straight line path. The amplitude of oscillation is 2 cm. When the displacement of the particle from the mean position is 1 cm, the numerical value of the magnitude of acceleration is equal to the numerical value of the magnitude of the velocity. The frequency of SHM is:

A
$$2\pi \sqrt{3}$$

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

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

D
$$\dfrac{1}{2\pi \sqrt{3}}$$

Solution

Let y is the displacement $$y=a\sin \omega t$$

$$\omega =\dfrac{2\pi }{T}$$

It is given that,

Displacement y = 1

Amplitude a = 2

So $$1=3\sin \dfrac{2\pi t}{T}$$

Since, acceleration in SHM is $$a={{\omega }^{2}}y$$

And velocity in SHM is $$v=\omega \sqrt{{{a}^{2}}-{{y}^{2}}}....(1)$$

$$\because v=\omega $$

Put all the values in equation (1)

$$ \dfrac{4{{\pi }^{2}}}{{{T}^{2}}}=\dfrac{2\pi }{T}\left( \sqrt{{{a}^{2}}-1} \right) $$

$$ \dfrac{2\pi }{T}=\left( \sqrt{3} \right) $$

$$ T=\dfrac{2\pi }{\sqrt{3}} $$
Question 7
1 / -0

The number of independent constituent simple harmonic motions yielding a resultant displacement equation of the periodic motion as $$y=8 sin^{2}(\frac{t}{2})sin (10t)$$ is:

A
8

B
6

C
4

D
3

Solution

The resultant displacement equation of periodic motion is given by

$$y=8sin^2(\dfrac{t}{2}). sin(10t)$$

$$y=4.2sin^2(\dfrac{t}{2}). sin (10t)$$

$$y=4[1-cost].sin(10t)$$ (as, $$2sin^2(\dfrac{t}{2})=1-cost$$)

$$y=4sin(10t)-4sin(10t).cost$$

$$y=4sin(10t)-2sin(11t)-sin(9t)$$ (as, $$2sinAtsinBt=sin(A+B)t+sin(A-B)t$$)

There are three independent constituent simple harmonic motion.

The correct option is D.
Question 8
1 / -0

A particle performs SHM with a period T and amplitude a. The mean velocity of the particle over thetime interval during which it travels a distance a/2 from the extreme position is

A
$$\dfrac{a}{T}$$

B
$$\dfrac{2a}{T}$$

C
$$\dfrac{3a}{T}$$

D
$$\dfrac{a}{2T}$$

Solution
Question 9
1 / -0

A particle is subjected to two mutually perpendicular simple harmonic motions such that its x andy coordinates are given by
$$x=2 \sin \omega t; y=2 \sin \left ( \omega t+\frac{\pi }{4} \right )$$ The path of the particle will be :

A
an ellipse

B
a straight line

C
a parabola

D
a circle

Solution
Question 10
1 / -0

A spring is suspended under gravity with a block attached to it. The block oscillates in the vertical plane according to $$x = a sin wt$$ with time period $$8$$ seconds. The ratio of velocities of the particle at $$t = 1$$ sec to $$t = 3$$ sec is

A
2 : 1

B
1 : 1

C
3 : 4

D
2 : 3

Solution

$$x = a sin \omega t$$
$$V = \dfrac {dx}{xt} = a\omega cos \omega t$$
$$T = \dfrac {2\pi}{\omega} ; \omega = \dfrac {2\pi}{T} = \dfrac {\pi}{4} \dfrac {rad}{sec}$$
$$\dfrac {V_1}{V_2} = \dfrac {cos \omega t_1}{cos \omega t_2} = \dfrac {cos \dfrac {\pi}{4}}{cos \dfrac {3\pi}{4}} = \dfrac {1}{1}$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Oscillations Test - 17

Result

Oscillations Test - 17

Score

-

Rank

-

SHARING IS CARING

Question 1

2 N

1.5 N

1 N

0.5 N

Solution

Question 2

$$30^{\circ}$$

$$45^{\circ}$$

$$60^{\circ}$$

$$90^{\circ}$$

Solution

Question 3

$$1$$ N

$$2$$N

$$3$$N

$$4$$N

Solution

Question 4

$$\dfrac{4\sqrt{2}}{3} \dfrac{A\omega }{\pi }$$

$$\dfrac{2A(4-\sqrt{2})}{3\pi }$$

$$\dfrac{3A\omega 4- \sqrt{2}\omega }{4\pi }$$

$$\dfrac{5\sqrt{2\pi }}{2\omega }$$

Question 5

$$-kx$$

$$-2kx$$

$$-\dfrac{2kx}{3}$$

$$\dfrac{4kx}{3}$$

Solution

Question 6

$$2\pi \sqrt{3}$$

$$\dfrac{2\pi }{\sqrt{3}}$$

$$\dfrac{\sqrt{3}}{2\pi }$$

$$\dfrac{1}{2\pi \sqrt{3}}$$

Solution

Question 7

8

6

4

3

Solution

Question 8

$$\dfrac{a}{T}$$

$$\dfrac{2a}{T}$$

$$\dfrac{3a}{T}$$

$$\dfrac{a}{2T}$$

Solution

Question 9

an ellipse

a straight line

a parabola

a circle

Solution

Question 10

2 : 1

1 : 1

3 : 4

2 : 3

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.