Waves Test

Result

Waves Test - 38

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

Two simple harmonic motions are given by:
$$ { x }_{ 1 }=a\sin { \omega t } +a\cos { \omega t } $$
$${ x }_{ 2 }=a\sin { \omega t } +\cfrac { a }{ \sqrt { 3 } } \cos { \omega t } $$
The ratio of the amplitudes of first and second motion and the phase difference between them are respectively:

A
$$\sqrt { \cfrac { 3 }{ 2 } } $$ and $$\cfrac { \pi }{ 12 } $$

B
$$\cfrac { \sqrt { 3 } }{ 2 } $$ and $$\cfrac { \pi }{ 12 } $$

C
$$\cfrac { 2 }{ \sqrt { 3 } } $$ and $$\cfrac { \pi }{ 12 } $$

D
$$\sqrt { \cfrac { 3 }{ 2 } } $$ and $$\cfrac { \pi }{ 6 } $$

Solution

$$\Rightarrow { x }_{ 1 }=a\sin { \omega t } +a\cos { \omega t } $$ $$ = $$ $$a\sqrt { 2 } \sin { (\omega t } +45)$$
$$\Rightarrow { x }_{ 2 }=a\sin { \omega t } +\cfrac { a }{ \sqrt { 3 } } \cos { \omega t } = \dfrac { 2a }{ \sqrt { 3 } } \sin ({ \omega t }+30)$$
Using these equations above phasor diagrams are drawn.
Ratio of amplitude $$=\cfrac { { a }_{ 1 } }{ { a }_{ 2 } } =\cfrac { \sqrt { 3 } }{ \sqrt { 2 } } $$
Phase difference $$=\cfrac { \pi }{ 4 } -\cfrac { \pi }{ 6 } =\cfrac { \pi }{ 12 } $$
Question 2
1 / -0

What is the phase difference between two simple harmonic motions represented by $$x_{1} = A\sin \left (\omega t + \dfrac {\pi}{6}\right )$$ and $$x_{2} = A \cos (\omega t)$$?

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

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

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

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

Solution

Given,
$$x_{1} = A \sin \left (\omega t + \dfrac {\pi}{6}\right )$$
$$x_{2} = A\cos (\omega t)$$
$$x_{2} = A \sin \left (\omega t + \dfrac {\pi}{2}\right )$$

Phase difference:
$$\triangle \phi = \phi_{2} - \phi_{1}$$

$$\triangle \phi = \dfrac {\pi}{2} - \dfrac {\pi}{6}$$

$$\triangle \phi = \dfrac {3\pi - \pi}{6} = \dfrac {\pi}{3}$$
Question 3
1 / -0

A pulse of a wave travelling on a string towards the fixed rigid end as shown in the figure above. The pulse is:

A
Reflected and transmitted

B
Reflected and refracted

C
Reflected and reduced

D
Reflected and magnified

E
Reflected and inverted

Solution

We know that , the wave undergoes a phase change of 180 degree on reflection from a rigid end.Therefore wave will be reflected and inverted.
Question 4
1 / -0

With the propagation of a longitudinal wave through a material medium, the quantities transmitted in the direction of propagation are

A
Energy, momentum and mass

B
Mass and momentum

C
Energy and mass

D
Energy and momentum

Solution

Hint:- Use properties of wave.
Explanation:-
$$\bullet$$ We know that during propagation of wave energy is transferred from one place to another without transfer of mass.
$$\bullet$$ Due to transfer in energy, momentum also transfer from one place to another,

Hence option D is correct
Question 5
1 / -0

Find out the amplitude of the travelling wave given in above figure?

A
$$0.08m$$

B
$$0.16m$$

C
$$0.32m$$

D
$$0.48m$$

E
$$0.60m$$

Solution

The peak to peak value of the wave $$=0.16 m$$
Thus, amplitude is the half of the peak to peak value i.e $$0.16/2=0.08$$ m
Question 6
1 / -0

The frequency of a sound wave is 200 Hz and its wavelength is 150 cm. What is the distance travelled by the sound wave in the time taken to produce 150 waves?

A
110 cm

B
225 cm

C
112.5 cm

D
336.5 cm

Solution

As frequency is $$200\,Hz$$, that means $$200$$ waves are produced on one second.

So, time taken to produce $$150$$ waves $$=\frac{150}{200}=0.75\,s$$

Velocity of wave = Frequency $$\times$$ Wavelength $$=200\times 1580=30000$$ cmps $$=300$$ meters per second

So, Distance $$=300\times 0.75=225\,cmeters$$
Question 7
1 / -0

Find out the change in wavelength of a $$700 Hz$$ acoustic wave as it enters brass from warm air? Given Speed of Sound in warm air is $$350 {m}/{s}$$ and through brass is $$3,500 {m}/{s}$$.

A
It decreases by a factor of $$20$$

B
It decreases by a factor of $$10$$

C
It increases by a factor of $$10$$

D
It increases by a factor of $$20$$

E
The wavelength remains unchanged when a wave passes into a new medium

Solution

Given : $$\nu = 700 $$ $$Hz$$ $$v_a = 350 m/s$$ $$v_b = 3500 m/s$$
Using $$v = \nu \lambda$$
$$\therefore$$ $$\lambda_a = \dfrac{v_a}{\nu} = \dfrac{350}{700} =0.5$$ $$m$$

Similarly $$\lambda_b = \dfrac{v_b}{\nu} = \dfrac{3500}{700} =5$$ $$m$$
Thus wavelength increases by a factor of $$10$$
Question 8
1 / -0

Two travelling waves of equal frequency, with amplitude $$4 cm$$ and $$6 cm$$ respectively, superimpose in a single medium. Find out the range of the amplitude, $$A$$, of the resultant wave?

A
$$2 cm \leq A \leq 10 cm$$

B
$$A = 5 cm$$

C
$$A = 10 cm$$

D
$$10 cm \leq A \leq 12 cm$$

E
$$12 cm \leq A \leq 24 cm$$

Solution

Given : $$A_1 = 4$$ cm, $$A_2 =6 cm$$
Maximum superimposition: The two waves interfere constructively.
Resultant amplitude $$A' = A_1 + A_2 = 4 + 6 =10 cm$$

Minimum superimposition: The two waves interfere destructively.
Resultant amplitude $$A'' = A_2 - A_1 = 6 - 4 =2 cm$$
Thus range of amplitude is $$2 cm \leq A \leq 10 cm$$
Question 9
1 / -0

Two simple harmonic motions are given by
$$x_1 = a \sin \omega t + a \cos \omega t$$ and
$$x_2 = a \sin \omega t + \dfrac{a}{\sqrt{3}}\cos \omega t$$
The ratio of the amplitudes of first and second motion and the phase difference between them are respectively

A
$$\sqrt{\dfrac{3}{2}}$$ and $$\dfrac{\pi}{12}$$

B
$${\dfrac{\sqrt{3}}{2}}$$ and $$\dfrac{\pi}{12}$$

C
$${\dfrac{3}{\sqrt{2}}}$$ and $$\dfrac{\pi}{12}$$

D
$$\sqrt{\dfrac{3}{2}}$$ and $$\dfrac{\pi}{6}$$

Solution

First SHM is given as $$x_1 = a\sin wt + a \cos wt = a [\sin wt + \cos wt]$$
$$\therefore$$ $$x_1 = \sqrt{2}a [\dfrac{1}{\sqrt{2}}\sin wt + \dfrac{1}{\sqrt{2}} \cos wt]$$
OR   $$x_1 = \sqrt{2}a [\cos(\pi /4)\sin wt + \sin{\pi/4} \cos wt]$$
$$\implies$$ $$x_1 = \sqrt{2} a \sin (wt + \pi/4)$$
Second SHM is given as $$x_2 = a\sin wt + \dfrac{a}{\sqrt{3}} \cos wt = a [\sin wt + \dfrac{1}{\sqrt{3}}\cos wt]$$
$$\therefore$$ $$x_2 = \dfrac{2}{\sqrt{3}}a [\dfrac{\sqrt{3}}{{2}}\sin wt + \dfrac{\sqrt{3}}{2\sqrt{3}} \cos wt]$$
OR   $$x_2 = \dfrac{2}{\sqrt{3}}a [\dfrac{\sqrt{3}}{{2}}\sin wt + \dfrac{1}{2} \cos wt]$$
OR   $$x_2 = \dfrac{2}{\sqrt{3}}a [\cos(\pi /6)\sin wt + \sin{\pi/6} \cos wt]$$
$$\implies$$ $$x_2 = \dfrac{2}{\sqrt{3}} a \sin (wt + \pi/6)$$
Thus ratio of magnitude $$ = \dfrac{\sqrt{2}a}{2a/\sqrt{3}} = \sqrt{\dfrac{3}{2}}$$
Also phase difference $$\Delta \phi = \dfrac{\pi}{4} - \dfrac{\pi}{6} = \dfrac{\pi}{12}$$
Question 10
1 / -0

The S.I unit of wavelength is ______________

A
$$cm$$

B
$$dm$$

C
$$m$$

D
$$km$$

Solution

wavelength is a measure of distance between to consecutive crest or trough so it's SI unit is m.
so the answer is C.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Waves Test - 38

Result

Waves Test - 38

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\sqrt { \cfrac { 3 }{ 2 } } $$ and $$\cfrac { \pi }{ 12 } $$

$$\cfrac { \sqrt { 3 } }{ 2 } $$ and $$\cfrac { \pi }{ 12 } $$

$$\cfrac { 2 }{ \sqrt { 3 } } $$ and $$\cfrac { \pi }{ 12 } $$

$$\sqrt { \cfrac { 3 }{ 2 } } $$ and $$\cfrac { \pi }{ 6 } $$

Solution

Question 2

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

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

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

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

Solution

Question 3

Reflected and transmitted

Reflected and refracted

Reflected and reduced

Reflected and magnified

Reflected and inverted

Solution

Question 4

Energy, momentum and mass

Mass and momentum

Energy and mass

Energy and momentum

Solution

Question 5

$$0.08m$$

$$0.16m$$

$$0.32m$$

$$0.48m$$

$$0.60m$$

Solution

Question 6

110 cm

225 cm

112.5 cm

336.5 cm

Solution

Question 7

It decreases by a factor of $$20$$

It decreases by a factor of $$10$$

It increases by a factor of $$10$$

It increases by a factor of $$20$$

The wavelength remains unchanged when a wave passes into a new medium

Solution

Question 8

$$2 cm \leq A \leq 10 cm$$

$$A = 5 cm$$

$$A = 10 cm$$

$$10 cm \leq A \leq 12 cm$$

$$12 cm \leq A \leq 24 cm$$

Solution

Question 9

$$\sqrt{\dfrac{3}{2}}$$ and $$\dfrac{\pi}{12}$$

$${\dfrac{\sqrt{3}}{2}}$$ and $$\dfrac{\pi}{12}$$

$${\dfrac{3}{\sqrt{2}}}$$ and $$\dfrac{\pi}{12}$$

$$\sqrt{\dfrac{3}{2}}$$ and $$\dfrac{\pi}{6}$$

Solution

Question 10

$$cm$$

$$dm$$

$$m$$

$$km$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday