Oscillations Test

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Oscillations Test - 41

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Question 1
1 / -0

A body of mass $$5\times 10^{-3}\ kg$$ is making $$S.H.M.$$ with amplitude $$1\times10 ^{-1}m$$ with maximum velocity $$1ms^{-1}$$, its velocity will be half at displacement of

A
$$5\times 10^{-3}m$$

B
$$2.5\times 10^{-3}m$$

C
$$5\sqrt{3}\times 10^{-2}m$$

D
$$\dfrac{5}{4}\times 10^{-3}m$$

Solution
Question 2
1 / -0

Which of the following equations do not represent a simple harmonic motion

A
$$ x= A cos (\omega t - \phi) + B sin (\omega t + \phi)$$

B
$$ x=B tan (\omega t + \phi)$$

C
$$ x= A e^{i(\omega t - \phi)}$$

D
$$x = Ae^{i \omega t} + B e^{i(\omega t -\phi)}$$

Solution
Question 3
1 / -0

A particle is performing simple harmonic motion having time period $$3s$$ is in phase with another particle which is also undergoing simple harmonic motion at $$t= 0$$ . The time period of second particle is T (less than 3s). If they are again in the same phase for the third time after $$45s$$, then the value of T is

A
$$2.8s$$

B
$$2.7s$$

C
$$2.5s$$

D
$$3.2s$$

Solution

Let $$\omega_1$$ and $$\omega_2$$ be the angular frequencies of first and second particle, respectively. Then, the phase by which they will proceed in time $$t$$ is $$\omega_1t$$ and $$\omega_2t$$, respectively.
According to the given situation,
$$\omega_2t-\omega_1t=3\times 2\pi$$ for $$t=45\ s$$

$$\dfrac{2\pi}{T}-\dfrac{2\pi}{3}=\dfrac{3\times 2\pi}{45}$$

$$\dfrac{1}{T}=\dfrac{1}{3}+\dfrac{1}{15}=\dfrac{6}{15}\Rightarrow T=2.5\ s$$
Question 4
1 / -0

A particle is executing S.H.M. between $$ x = \pm A. $$ The time taken to go from 0 to $$ \frac {A}{2} $$ is $$ T_1 $$ and to go from $$ \frac {A}{2} $$ to A is $$ T_2 $$ then:

A
$$ T_1 < T_2 $$

B
$$ T_1 >T_2 $$

C
$$ T_1 = T_2 $$

D
$$ T_1 = 2T_2 $$

Solution

let $$t=0, x=0$$
Then, $$x=A\sin (wt)$$
At $$t=T_1, x=\dfrac{A}{2}$$
$$\Rightarrow \dfrac{A}{2}=A\sin (wT_1)$$
$$\Rightarrow \sin wt_1=1/2$$
$$\Rightarrow T_1=\dfrac{\pi}{6w}$$
At $$t=(T_1+T_2), x=A$$
$$\Rightarrow A=A\sin (w[T_1+T_2])$$
$$\Rightarrow \sin [w(T_1+T_2)]=1$$
$$T_1+T_2=\dfrac{\pi}{2w}\Rightarrow T_2=\dfrac{\pi}{2w}-T_1$$
$$\Rightarrow T_2=\dfrac{\pi}{2w}-\dfrac{\pi}{6w}=\dfrac{\pi}{3w}$$
So, $$\dfrac{\pi}{3w}=2\left(\dfrac{\pi}{6w}\right)$$
$$\Rightarrow T_2=2T_1 \Rightarrow T_2>T_1$$
Option- $$A$$ is correct.
Question 5
1 / -0

A $$ 1.00 \times 10^{20} kg $$ particle is vibrating under simple harmonic motion with a period of $$ 1.00 \times 10^5 s $$ and with a maximum speed of $$ 1.00 \times 10^3 m/s $$ . The maximum displacement of particle from mean position is

A
$$1.59\ mm$$

B
$$1.00\ m$$

C
$$10\ m$$

D
$$None\ of\ these$$

Solution
Question 6
1 / -0

For a SHM with given angular frequency, two arbitrary initial conditions are necessary and sufficient to determine the motion completely.These initial conditions may be

A
Amplitude and initial phase

B
Amplitude and total energy of oscillation

C
Initial phase and total energy of oscillation

D
Initial position and initial velocity

Solution

Description of motion is completely specified if we know the variation of $$x$$ as a function of time. For a simple harmonic motion, the general equation of motion is $$x=A(\omega t+\delta )$$. As $$\omega$$ is given, to describe the motion completely, we need the values of $$A$$ and $$\delta$$.
For option (a), we can find $$A$$ and $$\delta $$ if we know initial velocity and initial position. Option (c) cannot give the values of $$A$$ and $$\delta$$ so it is not the correct condition.
Question 7
1 / -0

Springs of spring constants K, 2K, 4K, 8K, 2048 K are connected in series. A mass 'm' is attached to one end the system is allowed to oscillation. The time period is approximately :

A
$$2\pi \sqrt{\dfrac{m}{2K}}$$

B
$$2\pi \sqrt{\dfrac{2m}{2K}}$$

C
$$2\pi \sqrt{\dfrac{2m}{K}}$$

D
none of these

Solution

In series the equation spring constant is given by:-
$$\dfrac {1}{k'}=\displaystyle \sum \dfrac {1}{k'}\Rightarrow \dfrac {1}{k'}+\dfrac {1}{k}+\dfrac {1}{2k}+\dfrac {1}{4k}+\dfrac {1}{8k}+\dfrac {1}{2048k}$$
$$\Rightarrow \ \dfrac {1}{k'}=\dfrac {3841}{2048k}$$
or approximately $$k'\simeq \dfrac {2048k}{3841}\simeq \dfrac {k}{2}$$
$$\therefore \ $$ New time period, $$T'=2\pi \sqrt {\dfrac {m}{k'}}=2\pi \sqrt {\dfrac {2m}{k}}$$
Question 8
1 / -0

Two simple harmonic motion a shown below are at right angles. They are combined to form Lissajous figures. $$x ( t ) = A \sin ( a t + \delta )$$ $$y ( t ) = B \sin ( b t )$$ Identify the correct match below. Parameters curve

A
$$A \neq B , a = b ; \delta = 0$$ Parabola

B
$$A = B , a = b ; \delta = \pi / 2$$ line

C
$$A \neq B , a = b ; \delta = \pi / 2$$ Ellipse

D
$$A = B , a = 2 b ; \delta = \pi / 2$$ circle
Question 9
1 / -0

The bob of a $$0.2 m$$ pendulum describes an arc of circle in a vertical plane. If the tension in the cord is $$\sqrt { 3 }$$ times the weight of the bob when the cord makes an angle $$30$$ with the vertical, the acceleration of the bob in that position is

A
$$9.8$$

B
$$6.02$$

C
$$7.2$$

D
$$5.2$$

Solution
Question 10
1 / -0

A point mass oscillates along the $$x$$ -axis according to the law $$x = x _ { 0 } \cos ( \omega t - \pi / 4 )$$. If the acceleration of the particle is written as $$a = A \cos ( \omega t + \delta )$$ then:

A
$$A = x _ { 0 } , \delta = - \pi / 4$$

B
$$A = x _ { 0 } \omega ^ { 2 } , \delta = - \pi / 4$$

C
$$A = x _ { 0 } \omega ^ { 2 } , \delta = - \pi / 8$$

D
$$A = x _ { 0 } \omega ^ { 2 } , \delta = 3 \pi / 4$$

Solution

Given,
a particle oscillation about the law,
$$x=x_0\cos \left(wt-\dfrac{\pi}{4}\right)$$
and acceleration : $$a=A\cos (wt+\delta)$$
Since, maximum amplitude for SHM is given by :-
$$a_{max}=w^2.xmax$$ $$(w:$$ Angular frequency )
$$\Rightarrow A=x_0w^2$$
and, acceleration & displacement for an SHM are at opposite direction at any time $$t$$. So it results to a phase difference $$(\Delta \phi)$$ of $$\pi$$ between them.
$$\Rightarrow \Delta \phi =(wt+\delta)-(wt-\pi/4)=\pi$$
$$\Rightarrow \delta =\dfrac{3\pi}{4}$$.
so, eqn for acceleration :-
$$a=x_0w^2\cos \left(wt+\dfrac{3\pi}{4}\right)$$; $$A=x_0w^2, \delta =\dfrac{3\pi}{4}$$.

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Oscillations Test - 41

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Oscillations Test - 41

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Question 1

$$5\times 10^{-3}m$$

$$2.5\times 10^{-3}m$$

$$5\sqrt{3}\times 10^{-2}m$$

$$\dfrac{5}{4}\times 10^{-3}m$$

Solution

Question 2

$$ x= A cos (\omega t - \phi) + B sin (\omega t + \phi)$$

$$ x=B tan (\omega t + \phi)$$

$$ x= A e^{i(\omega t - \phi)}$$

$$x = Ae^{i \omega t} + B e^{i(\omega t -\phi)}$$

Solution

Question 3

$$2.8s$$

$$2.7s$$

$$2.5s$$

$$3.2s$$

Solution

Question 4

$$ T_1 < T_2 $$

$$ T_1 >T_2 $$

$$ T_1 = T_2 $$

$$ T_1 = 2T_2 $$

Solution

Question 5

$$1.59\ mm$$

$$1.00\ m$$

$$10\ m$$

$$None\ of\ these$$

Solution

Question 6

Amplitude and initial phase

Amplitude and total energy of oscillation

Initial phase and total energy of oscillation

Initial position and initial velocity

Solution

Question 7

$$2\pi \sqrt{\dfrac{m}{2K}}$$

$$2\pi \sqrt{\dfrac{2m}{2K}}$$

$$2\pi \sqrt{\dfrac{2m}{K}}$$

none of these

Solution

Question 8

$$A \neq B , a = b ; \delta = 0$$ Parabola

$$A = B , a = b ; \delta = \pi / 2$$ line

$$A \neq B , a = b ; \delta = \pi / 2$$ Ellipse

$$A = B , a = 2 b ; \delta = \pi / 2$$ circle

Question 9

$$9.8$$

$$6.02$$

$$7.2$$

$$5.2$$

Solution

Question 10

$$A = x _ { 0 } , \delta = - \pi / 4$$

$$A = x _ { 0 } \omega ^ { 2 } , \delta = - \pi / 4$$

$$A = x _ { 0 } \omega ^ { 2 } , \delta = - \pi / 8$$

$$A = x _ { 0 } \omega ^ { 2 } , \delta = 3 \pi / 4$$

Solution

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