Oscillations Test

Result

Oscillations Test - 38

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 body executing SHM at a displacement 'x' its PE is $$E_1$$, at a displacement 'Y' its PE is $$E_2$$ The P.E at a displacement $$(x+y)$$ is

A
$$\sqrt{E}=\sqrt{E_1}-\sqrt{E_2}$$

B
$$\sqrt{E}=\sqrt{E_1}+\sqrt{E_2}$$

C
$$E=E_1+E_2$$

D
$$E=E_1-E_2$$

Solution
Question 2
1 / -0

The vertical extension in a light spring by a weight of $$1\ kg$$ suspended from the wire is $$9.8\ cm$$. The period of oscillation:

A
$$ 20 \pi \,sec$$

B
$$ 2 \pi \,sec$$

C
$$\dfrac {2 \pi}{ \ 10}sec$$

D
$$ 200 \pi \,sec$$

Solution

$$\because mg=kx\Rightarrow \dfrac mk =\dfrac xg$$

$$\Rightarrow T=2\pi \sqrt{\dfrac{m}{k}}=2\pi \sqrt{\dfrac xg}$$

$$T=2\pi \sqrt{\dfrac{9.8\times 10^{-2}}{9.8}}=\dfrac{2\pi }{10}\sec$$
Question 3
1 / -0

A particle execute SHM with time period $$T$$ and amplitude $$A$$. The maximum possible average velocity in time $$\dfrac {T}{4}$$:

A
$$\dfrac {2A}{T}$$

B
$$\dfrac {4A}{T}$$

C
$$\dfrac {8A}{T}$$

D
$$\dfrac {4\sqrt {2}A}{T}$$

Solution

Given that,

Amplitude = $$A$$

Time period = $$T$$

Average velocity in time period = $$\dfrac{T}{4}$$

The displacement equation of SHM

$$x=A\sin \omega t$$

We know that,

$$ v=\dfrac{dx}{dt} $$

$$ v=\dfrac{d\left( A\sin \omega t \right)}{dt} $$

$$ v=A\omega \cos \omega t $$

Now, the average velocity is

$$ <v{{>}_{0\to \dfrac{T}{4}}}=\dfrac{\int\limits_{0}^{\dfrac{T}{4}}{vdt}}{\int\limits_{0}^{\dfrac{T}{4}}{dt}} $$

$$ <v>\,=\dfrac{\int\limits_{0}^{\frac{T}{4}}{A\omega \cos \omega t}}{\dfrac{T}{4}} $$

$$ <v>=\dfrac{4A\omega }{T}\int\limits_{0}^{\frac{T}{4}}{\cos \omega tdt} $$

$$ <v>=\dfrac{4A\omega }{T}\left[ \frac{\sin \omega t}{\omega } \right]_{0}^{\dfrac{T}{4}} $$

$$ <v>=\dfrac{4A}{T}\left[ \sin \dfrac{2\pi }{T}\times \dfrac{T}{4}-0 \right] $$

$$ <v>=\dfrac{4A}{T} $$

Hence, the average velocity is $$\dfrac{4A}{T}$$
Question 4
1 / -0

If y denotes the displacement and t denote the time and the displacement is given by $$y=a\,sin\,\omega t$$, the velocity of the particle is

A
$$ a\, cos\,\omega t$$

B
$$ -a\, cos\,\omega t$$

C
$$ a \omega\, cos\,\omega t$$

D
$$(a\,cos\,\omega t)/\omega$$

Solution
Question 5
1 / -0

The equation of S.H.M of a particle of amplitude 4 cm performing 150 oscillations per minute starting with an initial phase $$30^0$$ is

A
$$x=4 \, sin (2 \pi t+\pi/6)$$

B
$$x=4 \, sin (5 \pi t+\pi/6)$$

C
$$x=4 \, sin (2 \pi t+\pi/3)$$

D
$$x=4 \, sin (5 \pi t+\pi/3)$$

Solution
Question 6
1 / -0

A particle execute S.H.M from the mean position. its amplitude is $$A$$, its time period is "T'. At what displacement,its speed is half of its maximum speed.

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

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

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

D
$$\dfrac{3A}{\sqrt{A}}$$

Solution

For a SHM velocity, $$v=\omega \sqrt{A^2-x^2}$$. . . . . . .(1)
The maximum velocity is at $$x=0$$
$$v_m=\omega A$$
Now, $$v=\dfrac{\omega A}{2}$$

From equation (1),
$$\dfrac{\omega A}{2}=\omega \sqrt{A^2-x^2}$$
$$\dfrac{A^2}{4}=A^2-x^2$$
$$x^2=\dfrac{3A^2}{4}$$
$$x=\dfrac{\sqrt{3}A}{2}$$
The correct option is A.
Question 7
1 / -0

The displacement of a particle in S.H.M. is indicated by equation $$y\ = \ 10\ sin(20t\ +\ \pi/3)$$ where $$y$$ is in metres. The value of time period of vibration will be (in seconds):

A
$$ 10\pi sec$$

B
$$\pi /10 sec$$

C
$$10 sec$$

D
$$ 200sec$$

Solution
Question 8
1 / -0

A particle is oscillating according to the equation $$X = 7\cos 0.5\pi t$$, where $$t$$ is in seconds. The point moves from the position of equilibrium to maximum displacement in time.

A
$$4.0\ sec$$

B
$$2.0\ sec$$

C
$$1.0\ sec$$

D
$$0.5\ sec$$

Solution

From given equation $$\omega =\dfrac {2\pi}{T}=0.5 \pi \Rightarrow T=4$$ sec
Time taken from mean position to the maximum displacement $$=\dfrac {1}{4}T=1$$ sec.
Question 9
1 / -0

The maximum velocity of a particle performing linear S.H.M. is $$0.16\ m/s$$ and its maximum acceleration is $$0.64\ m/s^{2}$$. The calculate its time period

A
$$1.571\ s$$

B
$$0.08\ s$$

C
$$16\ s$$

D
$$3.2\ s$$

Solution

Given,
$$v_{max}=0.16m/s$$
$$a_{max}=0.64m/s^2$$
Time period, $$T=?$$

In S.H.M, we know that
$$v_{max}=A\omega$$. . . . . . .(1)
$$a_{max}=A\omega ^2$$. . . . . . (2)
$$\dfrac{a_{max}}{v_{max}}=\omega=\dfrac{2\pi}{T}$$

$$T=\dfrac{2\pi\times v_{max}}{a_{max}}$$

$$T=\dfrac{2\times 3.14\times0.16}{0.64}$$
$$T=1.57s$$
The correct option is A.
Question 10
1 / -0

Two particles are executing simple harmonic motion with same angular frequency but different amplitudes $$A$$ and $$2A$$ about same mean position on the same straight line. They cross each other when they are at a displacement of $$\dfrac{4A}{5}$$ from mean position in same direction. Then phase difference between them is:-

A
Zero

B
$$\cos^{-1}(\dfrac{1}{5}) -\cos^{-1}(\dfrac{2}{5})$$

C
$$53^\circ$$

D
$$\cos^{-1}(\dfrac{2}{5}) -\cos^{-1}(\dfrac{4}{5})$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Oscillations Test - 38

Result

Oscillations Test - 38

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\sqrt{E}=\sqrt{E_1}-\sqrt{E_2}$$

$$\sqrt{E}=\sqrt{E_1}+\sqrt{E_2}$$

$$E=E_1+E_2$$

$$E=E_1-E_2$$

Solution

Question 2

$$ 20 \pi \,sec$$

$$ 2 \pi \,sec$$

$$\dfrac {2 \pi}{ \ 10}sec$$

$$ 200 \pi \,sec$$

Solution

Question 3

$$\dfrac {2A}{T}$$

$$\dfrac {4A}{T}$$

$$\dfrac {8A}{T}$$

$$\dfrac {4\sqrt {2}A}{T}$$

Solution

Question 4

$$ a\, cos\,\omega t$$

$$ -a\, cos\,\omega t$$

$$ a \omega\, cos\,\omega t$$

$$(a\,cos\,\omega t)/\omega$$

Solution

Question 5

$$x=4 \, sin (2 \pi t+\pi/6)$$

$$x=4 \, sin (5 \pi t+\pi/6)$$

$$x=4 \, sin (2 \pi t+\pi/3)$$

$$x=4 \, sin (5 \pi t+\pi/3)$$

Solution

Question 6

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

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

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

$$\dfrac{3A}{\sqrt{A}}$$

Solution

Question 7

$$ 10\pi sec$$

$$\pi /10 sec$$

$$10 sec$$

$$ 200sec$$

Solution

Question 8

$$4.0\ sec$$

$$2.0\ sec$$

$$1.0\ sec$$

$$0.5\ sec$$

Solution

Question 9

$$1.571\ s$$

$$0.08\ s$$

$$16\ s$$

$$3.2\ s$$

Solution

Question 10

Zero

$$\cos^{-1}(\dfrac{1}{5}) -\cos^{-1}(\dfrac{2}{5})$$

$$53^\circ$$

$$\cos^{-1}(\dfrac{2}{5}) -\cos^{-1}(\dfrac{4}{5})$$

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.