Oscillations Test

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Oscillations Test - 43

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Weekly Quiz Competition

Question 1
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On suspending a mass M from a spring of force constant K,frequency of vibration f is obtained If a second spring as shown in the figure. is arranged then the frequency will be :

A
$$f\sqrt { 2 } $$

B
$$f/\sqrt { 2 } $$

C
2f

D
f

Solution

Given,

Time period of SHM is one spring and mass is connected

Spring constant, $$k$$

Attached mass, $$m$$

So, frequency is, $$f=2\pi \sqrt{\dfrac{k}{m}}$$

As in the figure, if another spring is added in parallel.

Both spring identical, new spring constant for two parallel spring.

$${{k}_{new}}=k+k=2k$$

Time period of oscillation,

$$ {{f}_{new}}=2\pi \sqrt{\dfrac{m}{{{k}_{new}}}}=2\pi \sqrt{\dfrac{m}{2k}} $$

$$ {{f}_{new}}=\dfrac{1}{\sqrt{2}}2\pi \sqrt{\dfrac{m}{k}}=\dfrac{f}{\sqrt{2}} $$

Hence, new frequency is $$\dfrac{f}{\sqrt{2}}$$
Question 2
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A particle executing $$SHM$$ while moving from the one extremity is found at distance $$x_1, x_2$$ and $$x_3$$ from the centre at the end of three successive seconds. The time period of oscillation is

A
$$\dfrac{2\pi}{\theta}$$

B
$$\dfrac{\pi}{\theta}$$

C
$$\theta$$

D
$$\dfrac{\pi}{2\theta}$$

Solution
Question 3
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The plot of velocity (v) versus displacement (x) of a particle executing simple harmonic motion is shown in figure. The time period of oscillation of particle is

A
$$\dfrac { \pi }{ 2 } s$$

B
$$\pi s$$

C
$$2\pi s$$

D
$$3\pi s$$

Solution

Row Hg
$$ V_{max} = Aw = 0.um/s$$
$$ x_{max} = A = 10 $$
$$ \frac{Aw}{A} = \frac{0.u}{10\times 10^{-2}} = \frac{40}{10} = 4 $$
$$ w = 4 $$
$$ T = \frac{2\pi }{w} = \frac{2\times \pi }{4}$$
$$ \therefore T = \pi /2^s$$
Answer is A
Question 4
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A simple pendulum has the time period of T at the surface of Earth. if it is taken to a height equal to R above the surface of Earth, then the new time period will be :

A
T

B
4T

C
$$\sqrt{2}T$$

D
2T

Solution

Option D is correct.
Question 5
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Figure shows the position-time graph of an object in SHM. The correct equation representing this motion is :

A
$$2 sin (\dfrac{2\pi }{5}t+\dfrac{\pi }{6})$$

B
$$4 sin (\dfrac{\pi }{5}t+\dfrac{\pi }{6})$$

C
$$4 sin (\dfrac{2\pi }{6}t+\frac{\pi }{3})$$

D
$$4 sin (\dfrac{2\pi }{6}t+\dfrac{\pi }{6})$$

Solution

$$T/2 = 11-5=6$$
Amplitude $$= A = 4$$
Time period = T = $$6\times 2= 12$$ second
So $$\omega =\dfrac{2\pi }{T}=\dfrac{2\pi }{12}$$
Let eq :- $$x = A\, \sin (\omega t+\phi )$$
$$x=y\, \sin (\dfrac{2\pi }{12}t+\phi )$$
at $$t=0, x=2$$
putting we get :-
$$\sin (\phi )= 1/2$$
So, $$\phi = \pi /6$$
So, eq :- $$x=4\, \sin \left(\dfrac{2\pi }{12}t+\dfrac{\pi }{6}\right)$$
option (d)
Question 6
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An object exceuting SHM has an acceleration of $$0.2m/s^{2}$$ at a distance of $$0.05$$m from the mean position calculate the period of SHM

A
$$2.1435$$

B
$$3.1428$$

C
$$1.432$$

D
$$4.321$$

Solution

Given,
$$a=0.2m/s^2$$
$$x=0.05m$$
The time period of S.H.M is given by
$$a=\omega ^2 x$$
$$a=(\dfrac{2\pi}{T})^2 x$$ ($$\omega =\dfrac{2\pi}{T}$$)
$$T^2=4\pi^2 \dfrac{x}{a}$$
$$T=2\pi \sqrt{\dfrac{x}{a}}$$
$$T=2\times 3.1428\times \sqrt{\dfrac{0.05}{0.2}}$$
$$T=3.1428 sec$$
The correct option is B.
Question 7
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The force of a required to row a boat at velocity is proportional to square of its speed of v km/ h requires 4 KW, how many does a seepd of 2V km/h required

A
8 kW

B
16 kW

C
32kw

D
76kw

Solution

Here $$F\propto V^{2}$$
$$P = F\times V$$
$$\propto V^{2}\times V = V^{3}$$
$$\therefore \dfrac{P_{1}}{P_{2}}=\dfrac{V_{1}^{3}}{V_{2}^{3}}=\left(\dfrac{V}{2V}\right)^{3}=\dfrac{1}{8}$$
$$\therefore P_{1}=4kw$$
$$P_{2}=8\times 4= 32kw$$
$$\therefore P_{2}= 32kw$$
Question 8
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The differential equation representing the SHM of a particule is $$\dfrac { { 9d }^{ 2 }y }{ { dt }^{ 2 } } $$ $$+ 4y = 0.$$ The time period of the particle is given by :

A
$$\dfrac\pi 3\ sec$$

B
$$\pi\ sec$$

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

D
$$3\pi\ sec$$

Solution

The differential equation representing S.H.M of a particle is,

$$9\dfrac{d^2y}{dt^2}+4y=0$$

$$9a+4y=0$$

$$a=-\dfrac{4}{9}y$$

$$\omega ^2=\dfrac{4}{9}$$ ($$a=-\omega ^2 y$$)

$$\omega =\sqrt{\dfrac{4}{9}}=\dfrac{2}{3}$$

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

$$T=3\pi sec$$

The correct option is D.
Question 9
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A particle executing $$S.H.M$$ of amplitude $$4\ cm$$ and $$T=4\ sec$$. The time taken by it move from positive extreme position to half the amplitude is:-

A
$$1\ sec$$

B
$$\dfrac {1}{3}\ sec$$

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

D
$$\sqrt{\dfrac {2}{3}}\ sec$$

Solution

If $$y$$ is the displacement, $$t$$ the time,$$a$$ the amplitude and $$\omega$$ the angular frequency
then $$y=a\cos{\omega t}$$
Given $$y=\dfrac{a}{2}\Rightarrow \dfrac{a}{2}=a\cos{\omega t}$$
$$\Rightarrow \cos{\omega t}=\dfrac{1}{2}$$
$$\Rightarrow \dfrac{1}{2}=\cos{\dfrac{2\pi}{T}\times 2}$$ at $$T=4$$secs
$$\Rightarrow t\cos{\dfrac{2\pi}{4}}=\dfrac{1}{2}$$
Let $${W}_{A}$$ be the work done by the spring $$A$$ and $${W}_{B}$$ be the work done by the spring $$B$$
$$\Rightarrow \dfrac{{W}_{A}}{{W}_{B}}=\dfrac{{K}_{A}}{{K}_{B}}=\dfrac{1}{3}$$
$$\therefore {W}_{A}:{W}_{B}=1:3$$
Question 10
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A particle is executing a simple harmonic motion. Its maximum acceleration is $$\alpha$$ and maximum velocity is $$\beta$$. Then, its time period of vibration will be

A
$$\dfrac {\beta^{2}}{\alpha}$$

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

C
$$\dfrac {\beta^{2}}{\alpha^{2 }}$$

D
$$\dfrac {\alpha}{\beta}$$

Solution