Waves Test

Result

Waves Test - 72

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The friction coefficient between the two blocks shown in the figure is $$\displaystyle \mu $$ and the horizontal plane is smooth. What can be the maximum amplitude $$(A)$$ if the upper block does not slip relative to the lower block ?

A
$$\displaystyle A=\frac{ (M+m)g}{k}$$

B
$$\displaystyle A=\frac{\mu (M+m)g}{k}$$

C
$$\displaystyle A=\frac{\mu (M+\mu m)g}{k}$$

D
$$\displaystyle A=\frac{\mu (\mu M+m)g}{k}$$

Solution
Question 2
1 / -0

A particle is executing SHM $$x=3\cos{\omega t} +4\sin{\omega t}$$. Find the phase shift and amplitude.

A
$$50^{\circ},5$$ units

B
$$37^{\circ},3.5$$ units

C
$$53^{\circ},3.5$$ units

D
$$37^{\circ},5$$ units

Solution

$$x=3\cos { \omega t } +4\sin { \omega t } $$
lets take
$$x_{ 0 }\cos \phi =4$$ ............$$(I)$$
and $$x_{ 0 }\sin \phi =3$$ ............$$(II)$$
Now the equation becomes
$$x=x_{ 0 }\sin \phi \cos { \omega t } +x_{ 0 }\cos \phi \sin { \omega t } \\ \Rightarrow x=x_{ 0 }\left( \sin \phi \cos { \omega t } +\cos \phi \sin { \omega t } \right) \\ \Rightarrow x=x_{ 0 }\sin { \left( \omega t+\phi \right) } $$
Squaring and adding $$(I)$$ and $$(II)$$ we get
$$\Rightarrow { x_{ 0 } }^{ 2 }(\cos ^{ 2 }{ \phi } +\sin ^{ 2 }{ \phi } )={ 4 }^{ 2 }+{ 3 }^{ 2 }=25\\ \Rightarrow x_{ 0 }=5$$
dividing $$(II)$$ by $$(I)$$, we get
$$\tan \phi =\frac { 3 }{ 4 } \\ \Rightarrow \phi =37^{ \circ }$$
$$\Rightarrow x=5\sin { \left( \omega t+37^{ \circ } \right) } $$ which represents a SHM of amplitude of 5 units and a phase of $$37^{ \circ }$$
Question 3
1 / -0

A person standing between the two vertical cliff produces a sound. Two successive echoes are heard at 4 s and 6 s. Calculate the distance between the cliffs :
(Speed of sound in air $$= 320 m s^{-1}$$)

A
1600 m

B
800 m

C
400 m

D
1200 m

Solution

Distance traveled by the sound from the $$1^{st}$$ cliff is calculated with the formula:
Distance traveled $$=$$ velocity of sound $$\times$$ time taken
That is, $$320 \times 4 $$ $$= 1280\ m$$.
This is twice the minimum distance between a source of sound (man) and the reflector ($$1^{st}$$ cliff) as it is reflected sound.
Distance traveled by the sound from the $$2^{nd}$$ cliff is calculated with the formula:
Distance traveled $$=$$ velocity of sound $$\times$$ time taken
That is, $$320 \times 6 = 1920\ m$$.
This is twice the minimum distance between a source of sound (man) and the reflector ($$2^{nd}$$ cliff) as it is reflected sound.
Hence, the distance between the 2 cliffs is given as: $$\dfrac { 1280 }{ 2 } +\dfrac { 1920 }{ 2 } =640+960=1600\ m$$
Question 4
1 / -0

Two wires of same material and area of cross section each of length 30 cm and 40 cm are stretched between two ends with tensions 10 N and 20 N respectively. The difference between the fundamental frequencies of two wires is 4.0 Hz, find the linear mass density of the wire.

A
$$16.4 \times 10^{-4} kg m^{-1}$$

B
$$16.4 \times 10^{-3} kg m^{-1}$$

C
$$6.4 \times 10^{-3} kg m^{-1}$$

D
$$6.4 \times 10^{-5} kg m^{-1}$$

Solution
Question 5
1 / -0

Directions For Questions

Two travelling waves of equal amplitudes and equal frequencies move in opposite direction along a string. They interfere to produce a standing wave having the equation $$\displaystyle y=A\: \cos \: kx\: \sin \: \omega t$$
in which A = 1.0 mm, $$\displaystyle k=1.57\: cm^{-1}$$ and $$\displaystyle \omega =78.5\: s^{-1}$$

...view full instructions

The velocity and amplitude of the component traveling waves are respectively

A
50 cm/s; 0.5 mm

B
50 cm/s; 5 mm

C
10 cm/s; 0.5 mm

D
50 cm/s; 1 mm

Solution

The standing wave is formed by the superposition of the waves
$$\displaystyle y_{1}=\frac{A}{2}\sin (\omega t-kx)$$ and $$\displaystyle y_{2}=\frac{A}{2}\sin (\omega t-kx)$$
The wave velocity (magnitude) of either of the waves is
$$\displaystyle v=\frac{\omega }{k}=\frac{78.5s^{-1}}{1.57cm^{-1}}=50$$ cm/s; Amplitude = 0.5 mm
Question 6
1 / -0

If for a particle moving in SHM, there is a sudden increase of $$1$$% in restoring force just as particle passing through mean position, percentage change in amplitude will be

A
$$1$$%

B
$$2$$%

C
$$0.5$$%

D
zero

Solution

As force in a simple harmonic motion is directly proportional to the square of amplitude.
$$F\propto { A }^{ 2 }$$
$$\dfrac { \triangle F }{ F } =2\dfrac { \triangle A }{ A } \\ \,\,\,\,\,\,\,\,\,\, =\dfrac { \triangle F }{ F } \times 100$$
$$\,\,\,\,\,\,\,\,\,\,\,=2\dfrac { \triangle A }{ A } \times 100\\ $$ % change in A.
% change in amplitude
$$\,\,\,\,\,\,=\dfrac { 1 }{ 2 } =0.5%$$
Question 7
1 / -0

A wave travelling along positive x-axis is given by $$=A\sin { \left( \omega t-kx \right) } $$. If it is reflected from a rigid boundary such that $$80$$% amplitude is reflected, then equation of reflected wave is

A
$$y=A\sin { \left( \omega t+0.8kx \right) } $$

B
$$y=-0.8A\sin { \left( \omega t+kx \right) } $$

C
$$y=A\sin { \left( \omega t+kx \right) } $$

D
$$y=0.8A\sin { \left( \omega t+kx \right) } $$

Solution

On getting reflected from a rigid boundary the wave suffers an additional phase change of $$\pi$$.
Hence, if $$\quad { y }_{ incident }=A\sin { \left( \omega t-kx \right) } $$
Then, $${ y }_{ reflected }=(0.8A)\sin { \left( \omega t-k(-x)+\pi \right) } =-0.8A\sin { \left( \omega t+kx \right) } $$
Question 8
1 / -0

Equations of a stationary wave and a travelling wave are $${ y }_{ 1 } = a\ sinkx\ cos \omega t$$ and $${ y }_{ 2 } = a\ sin (\omega t - kx)$$. The phase difference between two points $${ x }_{ 1 }\ =\ \dfrac { \pi }{ 3k } \ and\ { x }_{ 2 }\ =\ \dfrac { 3\pi }{ 2k } \ is\ { \phi }_{ 1 }$$ for the first wave and $${ \phi }_{ 2 }$$ for the second wave. The ratio $$\dfrac { { \phi }_{ 1 } }{ { \phi }_{ 2 } }$$ is :

A
$$1$$

B
$$\dfrac{5}{6}$$

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

D
$$\dfrac{6}{7}$$

Solution

Phase difference between two points in a standing wave $$=n\pi $$
where $$n$$ is the no of nodes between two points.
Given points are $${ x }_{ 1 }=\pi /3K=60/K$$

$${ x }_{ 2 }=3\pi /2K=210/K$$
Equation of the standing wave.
$${ y }_{ 1 }=asinKxcoswt$$
at node points, $$Kx=n\pi $$
$$x=\dfrac { n\pi }{ K } $$ $$(n=0,1,2,3......)$$
so nodes are $$=\dfrac { \pi }{ K } ,\dfrac { 2\pi }{ K } ,\dfrac { 3\pi }{ K } .....$$

$$\dfrac { 180 }{ K } ,\dfrac { 360 }{ K } ,\dfrac { 540 }{ K } $$

Since there is only one node in b/w phase difference $${ \phi }_{ 1 }=\pi $$
for travelling wave
$${ \phi }_{ 2 }=\dfrac { 2\pi }{ \lambda } \triangle x$$
from the equation, $${ y }_{ 2 }=a\sin\left( wt-Kx \right) $$
$$K=\dfrac { 2\pi }{ \lambda } $$
$$\therefore \quad { \phi }_{ 2 }=K\left[ { x }_{ 2 }-{ x }_{ 1 } \right] =K\left[ \dfrac { 3\pi }{ 2K } -\dfrac { \pi }{ 3K } \right] $$

$$=\pi \left[ \dfrac { 3 }{ 2 } -\dfrac { 1 }{ 3 } \right] =\dfrac { 7 }{ 6 } \pi $$

$$\therefore \dfrac{\phi_{1}}{\phi_{2}}=\dfrac{\pi}{\dfrac{7}{6}\pi}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{6}{7}$$
Question 9
1 / -0

A harmonic wave is travelling on string 1. At a junction with string 2, it is partly reflected and partly transmitted. The linear mass density of the second string is four times that of the first string and the boundary between the two strings is at x = 0. If the expression for the incident wave is $$\displaystyle y_{1}=A_{1}\: \cos (k_{1}x-\omega _{1}t)$$
What is the equation for the reflected wave in terms of $$\displaystyle A_{1},k_{1}$$ and $$\displaystyle \omega _{1}$$ ?

A
$$\displaystyle y =\frac{A_{1}}{4}\cos (k_{1}x+\omega _{1}t+\pi )$$

B
$$\displaystyle y =\frac{A_{1}}{6}\cos (k_{1}x+\omega _{1}t+\pi )$$

C
$$\displaystyle y =\frac{A_{1}}{3}\cos (k_{1}x+\omega _{1}t+\pi )$$

D
$$\displaystyle y =\frac{A_{1}}{3}\cos (2k_{1}x+\omega _{1}t+\pi )$$

Solution

Since $$\displaystyle v=\sqrt{T/\mu }$$ $$\displaystyle T_{2}=T_{1}\: \: and\: \: \mu _{2}=4\mu _{1}$$
we have $$\displaystyle v_{2}=\frac{v_{1}}{2}$$ ...........(i)
The frequency does not change that is
$$\displaystyle \omega _{1}=\omega _{2}$$ ................(ii)
Also because $$\displaystyle k=\omega /v$$ the wave numbers of the harmonic waves in the two strings are related by
$$\displaystyle k_{2}=\frac{\omega _{2}}{v_{2}}=\frac{\omega _{1}}{v_{1}/2}=2\frac{\omega _{1}}{v_{1}}=2k_{1}$$ ...................(iii)
The amplitudes are
$$\displaystyle A_{1}=\left ( \frac{2v_{2}}{v_{1}+v_{2}} \right )A_{1}=\left [ \frac{2(v_{1}/2)}{v_{1}+(v_{1}/2)} \right ]A_{1}=\frac{2}{3}A_{1}$$ .............(iv)
and $$\displaystyle A_{1}=\left ( \frac{v_{2}-v_{1}}{v_{1}+v_{2}} \right )A_{1}=\left [ \frac{(v_{1}/2)-v_{1}}{v_{1}+(v_{1}/2)} \right ]A_{1}=\frac{A_{1}}{3}$$ .................(v)
Now with equation (ii), (iii) and (iv) the transmitted wave can be written as
$$\displaystyle y_{1}=\frac{2}{3}A_{1}\cos (2k_{1}x-\omega _{1}t)$$ Ans
Similarly the reflected wave can be expressed as
$$\displaystyle =\frac{A_{1}}{3}\cos (k_{1}x+\omega _{1}t+\pi )$$
Question 10
1 / -0

A wave travels on a light string. The equation of the wave is $$Y = A\sin (kx - \omega t + 30^{\circ})$$. It is reflected from a heavy string tied to an end of the light string at $$x = 0$$. If $$64$$% of the incident energy is reflected the equation of the reflected wave is

A
$$Y = 0.8\ A\sin (kx - \omega t + 30^{\circ} + 180^{\circ})$$

B
$$Y = 0.8\ A\sin (kx + \omega t + 30^{\circ} + 180^{\circ})$$

C
$$Y = 0.8\ A\sin (kx + \omega t - 30^{\circ})$$

D
$$Y = 0.8\ A\sin (kx + \omega t + 30^{\circ})$$

Solution

$$y=A\sin\left( Kx-wt+{ 30 }^{ 0 } \right) $$
Intensity of the reflected wave $$=64$$%$$=0.64$$
Since $$I\alpha { A }^{ 2 }$$
Amplitude of the reflected wave $$=\sqrt { I } =\sqrt { 0.64 } $$
$$=0.8$$ of the original
i.e. $$80$$%
The reflected wave always have a phase difference of $$\pi $$
Therefore equation of the reflected wave is
$$\left[ y=0.8Asin\left( Kx+wt+{ 30 }^{ 0 }+{ 180 }^{ 0 } \right) \right] $$

Hence Option (B) is correct.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Waves Test - 72

Result

Waves Test - 72

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\displaystyle A=\frac{ (M+m)g}{k}$$

$$\displaystyle A=\frac{\mu (M+m)g}{k}$$

$$\displaystyle A=\frac{\mu (M+\mu m)g}{k}$$

$$\displaystyle A=\frac{\mu (\mu M+m)g}{k}$$

Solution

Question 2

$$50^{\circ},5$$ units

$$37^{\circ},3.5$$ units

$$53^{\circ},3.5$$ units

$$37^{\circ},5$$ units

Solution

Question 3

1600 m

800 m

400 m

1200 m

Solution

Question 4

$$16.4 \times 10^{-4} kg m^{-1}$$

$$16.4 \times 10^{-3} kg m^{-1}$$

$$6.4 \times 10^{-3} kg m^{-1}$$

$$6.4 \times 10^{-5} kg m^{-1}$$

Solution

Question 5

50 cm/s; 0.5 mm

50 cm/s; 5 mm

10 cm/s; 0.5 mm

50 cm/s; 1 mm

Solution

Question 6

$$1$$%

$$2$$%

$$0.5$$%

zero

Solution

Question 7

$$y=A\sin { \left( \omega t+0.8kx \right) } $$

$$y=-0.8A\sin { \left( \omega t+kx \right) } $$

$$y=A\sin { \left( \omega t+kx \right) } $$

$$y=0.8A\sin { \left( \omega t+kx \right) } $$

Solution

Question 8

$$1$$

$$\dfrac{5}{6}$$

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

$$\dfrac{6}{7}$$

Solution

Question 9

$$\displaystyle y =\frac{A_{1}}{4}\cos (k_{1}x+\omega _{1}t+\pi )$$

$$\displaystyle y =\frac{A_{1}}{6}\cos (k_{1}x+\omega _{1}t+\pi )$$

$$\displaystyle y =\frac{A_{1}}{3}\cos (k_{1}x+\omega _{1}t+\pi )$$

$$\displaystyle y =\frac{A_{1}}{3}\cos (2k_{1}x+\omega _{1}t+\pi )$$

Solution

Question 10

$$Y = 0.8\ A\sin (kx - \omega t + 30^{\circ} + 180^{\circ})$$

$$Y = 0.8\ A\sin (kx + \omega t + 30^{\circ} + 180^{\circ})$$

$$Y = 0.8\ A\sin (kx + \omega t - 30^{\circ})$$

$$Y = 0.8\ A\sin (kx + \omega t + 30^{\circ})$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.