Waves Test

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Waves Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

A simple harmonic oscillation is represented by the equation y $$= 0.4 sin \left ( \frac{440t}{7}+0.61x \right )$$. Where 'Y' and 'X' are in m and time 't' in seconds respectively. The value of time period in seconds

A
0.1

B
0.01

C
1

D
10

Solution

It is given that,

$$y=0.4\sin \left( \dfrac{440}{7}t+0.61x \right)$$

From the above equation:

$$ \omega =\dfrac{440}{7} $$

$$ \dfrac{2\pi }{T}=\dfrac{440}{7} $$

$$ \dfrac{2\times 22}{7T}=\dfrac{440}{7} $$

$$ T=0.1\,s $$
Question 2
1 / -0

If a body of mass 36gm moves with S.H.M of amplitude A =13 cm and period T= 12 sec. At a time t= ) the displacement is x = +13 cm.The shortest time of passage from x = +6.5 cm to x = -6.5 is:

A
4 sec

B
2 sec

C
6 sec

D
3 sec

Solution

The equation of the SHM is:-
$$y=A\cos \omega t$$ where $$ A=13cm$$ and $$\dfrac{2\pi}{T}$$
as $$T=125 , \Rightarrow \omega=\dfrac{2\pi }{12}=\dfrac{\pi}{6}rad/s$$
also at $$t=0,y=13$$ as given
$$=y=13 \cos \left(\dfrac{\pi t}{6}\right)$$ $$\Rightarrow t=\dfrac{6}{\pi}\cos^{-1}\left(\dfrac{y}{13}\right)$$
when particle is at $$y=6.5$$ ,time is,
$$t_1=\dfrac{6}{\pi}\cos^{-1}\left(\dfrac{6.5}{13}\right)=\dfrac{6}{\pi}\times \dfrac{\pi}{3}=25$$
also when $$y=-6.5$$, time is :-
$$t_2=\dfrac{6}{\pi}\cos^{-1}\left(\dfrac{-6.5}{13}\right)=\dfrac{6}{\pi}\times \dfrac{2\pi}{3}=45$$
$$\therefore $$ Time difference $$\Delta t=t_2-t_1=(4-2)=2s$$
Question 3
1 / -0

Three waveforms travelling along a straight line have the forms:
$$2A\sin \left (kx - \omega t + \dfrac {\pi}{3}\right ), \sqrt {3}A \cos \left (kx - \omega t - \dfrac {\pi}{3}\right ), 2\sqrt {3} \cos \left (kx - \omega t + \dfrac {\pi}{3}\right )$$. The amplitude of the resulting waveform is :

A
$$(2 + 3\sqrt {3})A$$

B
$$\sqrt {31}A$$

C
$$\sqrt {19}A$$

D
$$(2 - \sqrt {3})A$$
Question 4
1 / -0

Two waves having equation $$x _ { 1 } = a \sin \left( \omega t + \phi _ { 1 } \right)$$ and $$x _ { 2 } = a \sin \left( \omega t + \phi _ { 2 } \right)$$ If in the resultant wave the frequency and amplitude remains equals to amplitude of superimposing waves. Then phase difference between them-

A
$$\dfrac { \pi } { 6 }$$

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

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

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

Solution

$$x_{1}=a\, sin (\omega t+\phi _{1})$$
$$x_{2}= a\, sin (\omega t+\phi _{2})$$
$$\triangle \phi $$ = phase difference = $$\phi _{1}-\phi _{2}$$
Resulatant amplitude $$\Rightarrow a=\sqrt{(a)^{2}+(a)^{2}+2a^{2}cos \triangle \phi }$$
$$\Rightarrow a^{2}=2a^{2}+2a^{2}cos \triangle \phi $$
$$ cos \triangle \phi = -\frac{1}{2}$$
$$\triangle \phi = \frac{2\pi}{3}$$
Option B
Question 5
1 / -0

Three simple harmonic motions in the same direction having the same amplitude A and same period are superposed. If each differs in phase from the next by $$45^0$$, then

A
the resulting amplitude is $$\Big ( 1 + \sqrt{3} \Big ) A$$.

B
the resulting motion is not simple harmonic

C
The energy associated with the resulting motion ($$3 + 2 \sqrt{2}$$) times the energy associated with any single motion.

D
The phase of the resultant motion relative to the first is $$90^0$$

Solution

[C] $${Y_1} = A\sin \left( {wt - \dfrac{\pi }{4}} \right),\,\,{Y_2} = A\sin \left( {wt} \right),\,\,{Y_3} = A\left( {\sin wt + \dfrac{\pi }{4}} \right)$$
on superimposing, Resultant SHM =
$$\begin{array}{l} Y=A\left[ { 2\sin wt\cos \dfrac { \pi }{ 4 } +\sin wt } \right] \Rightarrow A\left[ { \sqrt { 2 } \sin wt+\sin wt } \right] \\ Y=A\left( { 1+\sqrt { 2 } } \right) \sin wt,\, \, { { Resultant } }\, \, AMP=\left( { 1+\sqrt { 2 } } \right) A \\ \frac { { \in { { resultant } } } }{ { \in { { single } } } } =\dfrac { { { { \left\{ { A\left( { 1+\sqrt { 2 } } \right) } \right\} }^{ 2 } } } }{ { { A^{ 2 } } } } \Rightarrow \in { { Resultant } }\, \, =\left( { 3+2\sqrt { 2 } } \right) { { single } } \end{array}$$
Question 6
1 / -0

Two waves $$E _ { 1 } = E _ { 0 } \sin \omega t$$ and $$E _ { 2 } = E _ { 0 } \sin ( \omega t + 60 )$$ superimpose each other. Find out initial phase of resultant wave?

A
$$30 ^ { \circ }$$

B
$$60 ^ { \circ }$$

C
$$120 ^ { \circ }$$

D
$$0 ^ { \circ }$$

Solution

Let the resultant wave be
$$E=E0' \sin (wt+\phi)$$./ Then
$$E=E_1+E_2$$
$$E=E_0\sin wt +E_0\sin (wt+60^o)$$
$$E=E_0(\sin wt +\sin (wt +60^o)$$
As $$\sin \cos +\sin (B)=2\sin \left (\dfrac {A+B}{2}\right) \cos \left (\dfrac {B-A}{2}\right) $$
$$\sin wt +\sin (wt +60^o)=2 \sin (wt +30^o).\cos (wt)$$
So, $$E=2E_0 \cos (wt) \sin (wt+30^o)$$
So, $$E_0=2E_0 \cos wt $$ and $$ \phi =30^o$$
Option $$A$$ is correct
Question 7
1 / -0

Two particles are executing simple harmonic motion of the same amplitude $$A$$ and frequency $$\omega$$ along the $$x-axis.$$ Their mean position is separated by distance $$X_0(X_0 > A)$$. If the maximum separation between them is $$(X_0 + A ),$$ the phase difference between their motion is :-

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

B
$$\dfrac{\pi}{6}$$

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

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

Solution

$${X_1} = A\sin \left( {wt + {\phi _1}} \right)$$
$${X_2} = A\sin \left( {wt + {\phi _2}} \right)$$
$${X_1} - {X_2} = A\left[ {2\sin \left[ {wt + \frac{{{\phi _1}}}{{{\phi _2}}}} \right]\sin \left[ {\frac{{{\phi _1} - {\phi _2}}}{2}} \right]} \right]$$
$$A = 2A\sin \left( {\frac{{{\phi _1} - {\phi _2}}}{2}} \right)$$
$$\frac{{{\phi _1} - {\phi _2}}}{2} = \frac{\pi }{6}$$
$${\phi _1} = \frac{\pi }{3}$$
Hence,
option $$(D)$$ is correct answer.
Question 8
1 / -0

Two $$SHMs$$ are given by $$Y_{1}= a\left[ \sin { \left( \dfrac { \pi }{ 2 } \right) } t+\varphi \right]$$ and $$Y_{2}= b\sin { \left[ \left( \dfrac { 2\pi t }{ 3 } \right) +\varphi \right] }$$ . The phase difference between these two after $$'1'\ sec$$ is:

A
$$\pi$$

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

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

D
$$\dfrac {\pi}{6}$$

Solution
Question 9
1 / -0

The distance between two consecutive crests in a wave train produced in string is 5 m. If two complete waves pass through any point per second, the velocity of wave is:

A
$$2.5 \mathrm { m } / \mathrm { s }$$

B
$$5 \mathrm { m } / \mathrm { s }$$

C
$$10 \mathrm { m } / \mathrm { s }$$

D
$$15 \mathrm { m } / \mathrm { s }$$

Solution

Hint:- Two wavelength passes through a point in given time so apply formula of velocity of wave accordingly.
Step 1: Note the Given values
Two consecutive crest =$$5\,m$$.
Two wave cross in time, $$t=\,1\sec $$
Wavelength = distance between two consecutive crest
$$\lambda =5\,m$$
Step 2: Calculate velocity of wave
\Velocity, $$v=\dfrac{2\lambda }{t}=\dfrac{2\times 5}{1}=10\,m{{s}^{-1}}$$
Hence, wave speed is $$10\,m{{s}^{-1}}$$
Question 10
1 / -0

A body is on a rough horizontal surface which is moving horizontally in $$SHM$$ of frequency $$2\ Hz$$. The coefficient of static friction between the body and the surface is $$0.5$$. The maximum value of the amplitude for which the body will not slip along the surface is approximately

A
$$9\ m$$

B
$$6\ m$$

C
$$4.5\ m$$

D
$$3\ m$$

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Re-Attempt Weekly Quiz Competition

Waves Test - 53

Result

Waves Test - 53

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Rank

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SHARING IS CARING

Question 1

0.1

0.01

1

10

Solution

Question 2

4 sec

2 sec

6 sec

3 sec

Solution

Question 3

$$(2 + 3\sqrt {3})A$$

$$\sqrt {31}A$$

$$\sqrt {19}A$$

$$(2 - \sqrt {3})A$$

Question 4

$$\dfrac { \pi } { 6 }$$

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

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

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

Solution

Question 5

the resulting amplitude is $$\Big ( 1 + \sqrt{3} \Big ) A$$.

the resulting motion is not simple harmonic

The energy associated with the resulting motion ($$3 + 2 \sqrt{2}$$) times the energy associated with any single motion.

The phase of the resultant motion relative to the first is $$90^0$$

Solution

Question 6

$$30 ^ { \circ }$$

$$60 ^ { \circ }$$

$$120 ^ { \circ }$$

$$0 ^ { \circ }$$

Solution

Question 7

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

$$\dfrac{\pi}{6}$$

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

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

Solution

Question 8

$$\pi$$

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

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

$$\dfrac {\pi}{6}$$

Solution

Question 9

$$2.5 \mathrm { m } / \mathrm { s }$$

$$5 \mathrm { m } / \mathrm { s }$$

$$10 \mathrm { m } / \mathrm { s }$$

$$15 \mathrm { m } / \mathrm { s }$$

Solution

Question 10

$$9\ m$$

$$6\ m$$

$$4.5\ m$$

$$3\ m$$

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