Oscillations Test

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Oscillations Test - 48

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Weekly Quiz Competition

Question 1
1 / -0

A particle is executing S.H.M.with amplitude $$5$$ cm along x axis, origin as mean position. If at $$x= +4$$ cm magnitude of velocity is equal to magnitude of acceleration then find time period of oscillation in seconds.

A
$$\frac{4\pi}{3}$$

B
$$6\pi$$

C
$$\frac{8\pi}{3}$$

D
$$\frac{9\pi}{2}$$

Solution
Question 2
1 / -0

A simple harmonic oscillator of angular frequency $$2$$ rad/s is acted upon by an external force $$F = \sin t$$ N. If the oscillator is at rest in its equilibrium position at $$t= 0$$, its position at later times is proportional to:

A
$$ \sin t + \cfrac{1}{2} \cos 2t$$

B
$$ \cos t - \cfrac{1}{2} \sin 2t$$

C
$$ \sin t + \cfrac{1}{2} \sin 2t$$

D
$$ \sin t - \cfrac{1}{2} \sin 2t$$

Solution

According to the question,
Given; $$F= Fsin t \Rightarrow a = \dfrac{F}{m} sin t \Rightarrow v = \dfrac{-F}{m} cos t$$
$$x = \dfrac{-F}{m} sin t \Rightarrow X = r \vec{x_1} + r \vec{x_2} = Asin 2t - \dfrac{F}{m} sint$$..............................$$eq-i$$

Now, we know that in $$SHM$$,
$$a \ ( acceleration) = - \omega^2 A sin \omega t$$ and above we find $$a = \dfrac{F}{m} sin t $$

Resultant force, $$F_{resultant} = sint - m \omega^2 A sinwt$$

$$\Rightarrow F_R = sint - 4mAsin2t$$ (Given, $$\omega = 2 radian/s$$)
$$\Rightarrow a_R = sint /m- 4Asin2t$$ (By integrating)
$$\Rightarrow v_R = -cost /m + 2Acos2t$$ (By integrating)
$$\Rightarrow x_R = -sint /m + Asin2t$$

We can write, $$x_R = \dfrac{-1}{m} (sint - Amsin2t) \ \Rightarrow x_R \propto (sint - Amsin2t)$$

Only option $$D$$ is in the given form, Hence, option $$D$$ is correct.
Question 3
1 / -0

If amplitude of particle executing $$SHM$$ is doubled, which of the following quantities are doubled
i) Time period
ii) Maximum velocity
iii) Maximum acceleration
iv) Total energy

A
ii & iii

B
i, ii & iii

C
i, & iii

D
i, ii, iii & iv
Question 4
1 / -0

A simple harmonic motion has an amplitude A and time period T. Find the time required by it to travel diameter from .

A
$$x = x\,to\,x = \,A/2$$

B
$$x\, = \,0\,\,to\,x\, = \,\frac{A}{{\sqrt 2 }}$$

C
$$x = A\,to\,x = \,A/2$$

D
$$x = - \frac{A}{{\sqrt 2 }}\,to\,x = \,\frac{A}{{\sqrt 2 }}$$
Question 5
1 / -0

A simple motion is represented by:
$$y = 5(\sin 3 \pi t + \sqrt{3} \cos 3\pi t)cm$$
The amplitude and time period of the motion are:

A
$$5 \,cm, \dfrac{3}{2}s$$

B
$$5 \,cm, \dfrac{2}{3} s$$

C
$$10 \,cm, \dfrac{3}{2}s$$

D
$$10 \,cm, \dfrac{2}{3}s$$

Solution

$$y = 5[\sin (3\pi t) + \sqrt{3} \cos (3\pi t)]$$

$$= 10 \sin \left(3\pi t + \dfrac{\pi}{3}\right)$$

Amplitude $$= 10 \,cm$$

$$T = \dfrac{2\pi}{w} = \dfrac{2\pi}{3\pi} = \dfrac{2}{3} \sec$$
Question 6
1 / -0

A body oscillates with SHM according to the equation $$x=(5.0\quad m)\cos { [(2\pi \quad rad\quad { s }^{ -1 } } )t+{ \pi }/{ 4] }$$
At t=1.5 s, its acceleration is:

A
$$-139.56\quad { m }/{ { s }^{ 2 } }$$

B
$$139.56\quad { m }/{ { s }^{ 2 } }$$

C
$$69.78\quad { m }/{ { s }^{ 2 } }$$

D
$$-69.78\quad { m }/{ { s }^{ 2 } }$$

Solution

$$x=5\cos \left(2\pi t+\dfrac{\pi}{4}\right)$$
$$v=\dfrac{dx}{dt}=-5(2\pi)\sin \left(2\pi t+\dfrac{\pi}{4}\right)$$
$$a=\dfrac{d^2x}{dt^2}=-5(2\pi)^2\cos \left(2\pi t+\dfrac{\pi}{4}\right)$$
Put $$t=1.5$$
$$a=+139.56m/s^2$$
$$\cos \left(3\pi +\dfrac{\pi}{4}\right)=\dfrac{-1}{\sqrt{2}}$$
So option (B).
Question 7
1 / -0

Which of the following quantity is unitless

A
Velocity gradient

B
Pressure gradient

C
Displacement gradient

D
NONE

Solution
Question 8
1 / -0

A particle executes simple harmonic motion with an amplitude of 5 cm. when the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. then, its periodic time in second is:

A
$$\dfrac{7}{3}\pi$$

B
$$\dfrac{3}{8}\pi$$

C
$$\dfrac{4\pi}{3}$$

D
$$\dfrac{8\pi}{3}$$

Solution

We know that velocity in S. H. M. is given by
$$v=\omega \sqrt{a^2-x^2}$$
$$v=\omega \sqrt{5^2-4^2}=3\omega$$
$$a=\omega^2\times 4$$
Given |A| = |v|
$$4\omega^2=3\omega$$
$$\omega =\dfrac{3}{4}=\dfrac{2H}{T}$$
$$T=\dfrac{8H}{3}$$ sec.
Question 9
1 / -0

After charging a capacitor $$C$$ to a potential $$V$$ , it is connected across an ideal inductor $$L$$.The capacitor starts discharging simple harmonically at time $$t = 0 .$$The charge on the capacitor at a later time instant is $$q$$ and the periodic time of simple harmonic oscillations is $$T$$. Therefore,

A
$$q = C V \sin ( \omega t )$$

B
$$q = C V \cos ( \omega t )$$

C
$$T = 2 \pi \sqrt { \frac { 1 } { L C } }$$

D
$$T = 2 \pi \sqrt { L C }$$
Question 10
1 / -0

Two particles undergoing simple harmonic motion of same frequency and same amplitude cross each other at $$x=\dfrac {A}{2}$$. Phase difference between them is

A
$$\dfrac {\pi}{3}$$

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

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

D
$$\dfrac {\pi}{4}$$

Solution

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Oscillations Test - 48

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Oscillations Test - 48

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SHARING IS CARING

Question 1

$$\frac{4\pi}{3}$$

$$6\pi$$

$$\frac{8\pi}{3}$$

$$\frac{9\pi}{2}$$

Solution

Question 2

$$ \sin t + \cfrac{1}{2} \cos 2t$$

$$ \cos t - \cfrac{1}{2} \sin 2t$$

$$ \sin t + \cfrac{1}{2} \sin 2t$$

$$ \sin t - \cfrac{1}{2} \sin 2t$$

Solution

Question 3

ii & iii

i, ii & iii

i, & iii

i, ii, iii & iv

Question 4

$$x = x\,to\,x = \,A/2$$

$$x\, = \,0\,\,to\,x\, = \,\frac{A}{{\sqrt 2 }}$$

$$x = A\,to\,x = \,A/2$$

$$x = - \frac{A}{{\sqrt 2 }}\,to\,x = \,\frac{A}{{\sqrt 2 }}$$

Question 5

$$5 \,cm, \dfrac{3}{2}s$$

$$5 \,cm, \dfrac{2}{3} s$$

$$10 \,cm, \dfrac{3}{2}s$$

$$10 \,cm, \dfrac{2}{3}s$$

Solution

Question 6

$$-139.56\quad { m }/{ { s }^{ 2 } }$$

$$139.56\quad { m }/{ { s }^{ 2 } }$$

$$69.78\quad { m }/{ { s }^{ 2 } }$$

$$-69.78\quad { m }/{ { s }^{ 2 } }$$

Solution

Question 7

Velocity gradient

Pressure gradient

Displacement gradient

NONE

Solution

Question 8

$$\dfrac{7}{3}\pi$$

$$\dfrac{3}{8}\pi$$

$$\dfrac{4\pi}{3}$$

$$\dfrac{8\pi}{3}$$

Solution

Question 9

$$q = C V \sin ( \omega t )$$

$$q = C V \cos ( \omega t )$$

$$T = 2 \pi \sqrt { \frac { 1 } { L C } }$$

$$T = 2 \pi \sqrt { L C }$$

Question 10

$$\dfrac {\pi}{3}$$

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

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

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

Solution

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