Oscillations Test

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Oscillations Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

The equation of a progressive wave can be given by $$y = 15 \sin (660 \pi t- 0.02 \pi x) cm$$. The frequency of the wave is :

A
$$330 Hz$$

B
$$342 Hz$$

C
$$365 Hz$$

D
$$660 Hz$$

Solution

$$y = 15$$ sin $$(660 \pi t - 0.02 \pi x)$$
The general equation of a progressive wave is
$$y(x,t)= A \sin (\omega t - kx)$$
$$\omega=2 \pi \nu$$
$$\nu= \dfrac{660 \pi} {2 \pi}$$
=330 Hz
Question 2
1 / -0

A simple pendulum executing $$SHM$$ with a period of $$6s$$ between two extreme positions $$B$$ and $$C$$ about a point $$O$$. If the length of the arc $$BC$$ is $$10$$cm, how long will the pendulum take the move from position $$C$$ to a position $$D$$ towards $$O$$ exactly midway between $$C$$ and $$O$$?

A
$$0.5s$$

B
$$1s$$

C
$$1.5s$$

D
$$3s$$

Solution

O is the mean position and B and C are extreme positions.
Given: BC = $$10$$cm
Since B and C are the extreme positions, therefore amplitude of the SHM oscillation is 5cm
Now time taken for covering 10cm =6s
so time for $$10/4$$cm =$$\dfrac{6*2.5}{10}$$
=1.5s
D is the midway between C and O.
Question 3
1 / -0

The displacement of a particle varies with time according to the relation $$y = a sin \omega t + b cos \omega t.$$

A
The motion is oscillatory but not SHM.

B
The motion is SHM with amplitude a + b.

C
The motion is SHM with amplitude $$a^2 + b^2$$

D
The motion is SHM with amplitude $$\sqrt{a^2 + b^2}$$

Solution

Given $$x = asin \omega t + b cos \omega t$$ .......(i)
Let $$a = A cos \phi$$ ........(ii)
and $$b A sin \phi$$ ..........(iii)
Squaring adding (ii) and (iii), we get
$$a^2 + b^2 = A^2 cos^2 \phi + A^2 sin^2 \phi = A^2$$
Eq.(i) can be written as
$$x = A cos \phi sin \omega t + A sin \phi cos \omega t = A sin (\omega t + \phi)$$
It is equation of SHM with amplitude $$A = \sqrt{a^2 + b^2}$$
Question 4
1 / -0

The amplitude of a particle in SHM is $${\text{5 }}cms$$ and its time period is $$\pi .$$ At a displacement of $$3\;cm$$ from its mean position the velocity in cm/sec will be:

A
8

B
12

C
2

D
16

Solution

Let equation of SHM is $$x= Asin(\omega t)$$, Where $$A= 5 $$ $$cms$$,
$$\therefore v= A\omega cos(\omega t)$$
Given, Time Period$$=\pi= \dfrac{2\pi}{\omega}$$ $$\Rightarrow \omega= 2$$
$$x= 5 sin (2t)$$,
At a displacement $$3$$ $$cm$$, $$\Rightarrow 3= 5 sin(2t)$$ $$\Rightarrow cos(2t) = \dfrac{4}{5}$$,
$$v= 5\times 2cos(2t)= 5\times 2\times \dfrac{4}{5}=8 cms^{-1}$$
Question 5
1 / -0

The force on a particle of mass $$10\ g$$ is $$(10\hat{i}+5\hat{j})N$$. If it starts from rest, what would be its position at time $$t = 5 s$$?

A
$$(12500\hat{i}+6250\hat{j})m$$

B
$$(6250\hat{i}+12500\hat{j})m$$

C
$$(12500\hat{i}+12500\hat{j})m$$

D
$$(6250\hat{i}+6250\hat{j})m$$

Solution

Mass of particle $$=10g=10\times { 10 }^{ -3 }Kg$$
$$F=10\hat { i } +5\hat { j } $$
$$Ma=10\hat { i } +5\hat { j } $$
$$\dfrac { 10 }{ 1000 } a=10\hat { i } +5\hat { j } $$
$$\Rightarrow a=1000\hat { i } +500\hat { j } $$
$$\Rightarrow \dfrac { dV }{ dt } =1000\hat { i } +500\hat { j } $$
$$\Rightarrow dV=1000dt\hat { i } +500dt\hat { j } $$
$$\Rightarrow V=1000t\hat { i } +500t\hat { j } $$
$$\Rightarrow \dfrac { dx }{ dt } =100t$$ and $$\dfrac { dy }{ dt } =500t$$
$$\Rightarrow \int { dx } =\int _{ 0 }^{ 5 }{ 100t } dt\quad \quad \int { dy } =\int _{ 0 }^{ 5 }{ 500t } dt$$
$$x=\dfrac { 100{ \left( 5 \right) }^{ 2 } }{ 2 } \hat { i } \quad \quad y=\dfrac { 500{ \left( 5 \right) }^{ 2 } }{ 2 } \hat { j } $$
$$x=12500\hat { i } \quad \quad \quad y=6250\hat { j } $$
position $$=12500\hat { i } +6250\hat { j } $$

So, the correct answer is 'A'.
Question 6
1 / -0

An object with a mass M is suspended from an elastic spring with a spring constant k. The object oscillates with period T. If the mass of oscillations is quadrupled, how it will change the period of oscillations?

A
The period is increased by factor two

B
The period is increased by factor four

C
The period is decreased by factor two

D
The period remains the same

Solution

The time period of oscillations is given by the formula $$T = 2\pi \sqrt(\frac{M}{K})$$
If the mass is quadrupled (i.e 4M), then time period becomes doubled
Question 7
1 / -0

Two pendulums differ in lengths by $$22\ cm$$. They oscillate at the same place so that one of them makes $$30$$ oscillations and the other makes $$36$$ oscillations during the same time. The lengths (in cm) of the pendulums are

A
$$72$$ and $$50$$

B
$$60$$ and $$38$$

C
$$50$$ and $$28$$

D
$$80$$ and $$58$$

Solution

Given,
$$T_1=\dfrac{1}{30}$$
$$T_2=\dfrac{1}{36}$$
The time period of the pendulum is given by
$$T=2\pi \sqrt{\dfrac{l}{g}}$$
$$T\propto \sqrt{l}$$
$$\dfrac{T_1}{T_2}=\sqrt{\dfrac{l_1}{l_2}}=\dfrac{36}{30}$$
$$\sqrt{\dfrac{l_1}{l_2}}=\dfrac{6}{5}$$
$$\dfrac{l_1}{l_2}=\dfrac{36}{25}$$. . . . . .(1)
Accorrding to question,
$$\dfrac{11}{25}=\dfrac{l_1-l_2}{l_2}=\dfrac{22}{l_2}$$. . . . .(2)
From equation (1) and (2), we get
$$l_2=50cm$$
$$l_1=72cm$$
The correct option is A.
Question 8
1 / -0

A particle executes SHM with an amplitude of $$2cm$$. When the particle is at $$1cm$$ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

A
$$\cfrac { 1 }{ 2\pi \sqrt { 3 } } $$

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

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

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

Solution

Given,
$$A=2cm$$
$$x=1cm$$
The relation of velocity and displacement is given by
$$v=\omega \sqrt{A^2-x^2}$$
Acceleration, $$a=-\omega ^2x$$
According to question,
$$\omega ^2 x=\omega \sqrt{A^2-x^2}$$
$$\omega x=\sqrt{A^2-x^2}$$
$$\dfrac{2\pi}{T}\times 1=\sqrt{2^2-1^2}=\sqrt{4-1}$$
$$T=\dfrac{2\pi}{\sqrt{3}}$$
The correct option is C.
Question 9
1 / -0

The maximum velocity of a particle executing simple harmonic motion is $$v$$. If the amplitude is doubled and the time period of oscillation decreased to $$1/3$$ of its original value, the maximum velocity becomes:

A
$$18v$$

B
$$12v$$

C
$$6v$$

D
$$3v$$

Solution

Lets consider $$v=$$the maximum velocity of a particle executing harmonic motion
$$v=A\omega$$. . . . . . . .(1)
After the time decreased $$1/3$$ of its original value,
$$T'=\dfrac{1}{3}T$$
$$A'=2A$$
$$\omega '=\dfrac{2\pi}{T'}=3\omega$$
$$v '=A'\omega '$$
$$v'=2A\times 3\omega$$
$$v'=6A\omega =6v$$ (from equation 1)
The correct option is C.
Question 10
1 / -0

Motion of an oscillating liquid in a U tube is

A
periodic but not simple harmonic

B
non - periodic

C
simple harmonic and time period is independent of the density of the liquid.

D
simple harmonic and time period is directly proportional to the density of the liquid.

Solution

Motion of an oscillating liquid column in a U tube is SHM with period, $$T = 2\pi \sqrt{\frac{h}{g}}$$, where h is the height of liquid column in one arm of U tube in equilibrium position of liquid. Therefore, T is independent of density of liquid.

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Re-Attempt Weekly Quiz Competition

Oscillations Test - 32

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Oscillations Test - 32

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Question 1

$$330 Hz$$

$$342 Hz$$

$$365 Hz$$

$$660 Hz$$

Solution

Question 2

$$0.5s$$

$$1s$$

$$1.5s$$

$$3s$$

Solution

Question 3

The motion is oscillatory but not SHM.

The motion is SHM with amplitude a + b.

The motion is SHM with amplitude $$a^2 + b^2$$

The motion is SHM with amplitude $$\sqrt{a^2 + b^2}$$

Solution

Question 4

8

12

2

16

Solution

Question 5

$$(12500\hat{i}+6250\hat{j})m$$

$$(6250\hat{i}+12500\hat{j})m$$

$$(12500\hat{i}+12500\hat{j})m$$

$$(6250\hat{i}+6250\hat{j})m$$

Solution

Question 6

The period is increased by factor two

The period is increased by factor four

The period is decreased by factor two

The period remains the same

Solution

Question 7

$$72$$ and $$50$$

$$60$$ and $$38$$

$$50$$ and $$28$$

$$80$$ and $$58$$

Solution

Question 8

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

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

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

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

Solution

Question 9

$$18v$$

$$12v$$

$$6v$$

$$3v$$

Solution

Question 10

periodic but not simple harmonic

non - periodic

simple harmonic and time period is independent of the density of the liquid.

simple harmonic and time period is directly proportional to the density of the liquid.

Solution

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