Waves Test

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Waves Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

If $${y}_{1}=5(mm)]sin{\pi t}$$ and that of $${S}_{2}$$ is $${y}_{2}=5(mm)\sin{(\pi t+\pi /6)}$$. A wave from $${S}_{1}$$ reaches point A in $$1\ sec$$ while a wave from $${S}_{2}$$ reaches point A in $$0.5s$$. The resulting amplitude at point $$A$$ is :-

A
$$5\sqrt { 2+\sqrt { 3 } } mm$$

B
$$5\sqrt { 3 } mm$$

C
$$5mm$$

D
$$5\sqrt { 2 }mm $$

Solution

$$Y_{1}=5 (mm) \sin \pi t$$
$$Y_{2}=5 (mm)\sin \left(\pi t+\dfrac{\pi}{6}\right)$$
from both transverse wave equation
we can get
$$A_{1}=5\ mm$$
$$A_{2}=5\ mm$$
$$\phi=(\pi t)+\pi/6-\pi t)=\dfrac{\pi}{6}$$
where, $$A_{1}$$ and $$A_{2}$$ are amplitude of wave equation and $$\phi$$ is the phase difference
resultant amplitude$$=\sqrt{A_{1}^{2}+A_{2}^{2}+2A_{1}A_{2}\cos\phi}$$
$$=5\sqrt{1+1+2\times 1\times 1\times \dfrac{\sqrt{3}}{2}}$$
$$=5\sqrt{2+\sqrt{3}}$$
Resultant Amplitude $$=5\sqrt{2+\sqrt{3}}mm$$
Correct option is $$(A)$$
Question 2
1 / -0

Displacement time equation of a particle executing SHM is x=4 sin$$\omega$$t + 3 sin ($$\omega t+ \dfrac{\pi}{3})$$ here x is in centimeter and t is in seconds . The amplitude of oscillation of the particle is approximately:

A
5 cm

B
6 cm

C
7 cm

D
9 cm

Solution

Resultant amplitude of given expression is A.
$$A=\sqrt{A_1^2+A_2^2+2A_1A_2\cos\phi}$$
$$=\sqrt{4^2+3^2+2\times 4\times 3\cos \dfrac{\pi}{3}}$$
$$=\sqrt{16+9+12}=\sqrt{37} \cong6cm$$
Question 3
1 / -0

Two waves of frequencies 20 Hz and 30 Hz travels out from a common point. The phase difference between them after 0.6 sec is

A
$$12\pi $$

B
$${\pi \over 2}$$

C
$$\pi $$

D
$${{3\pi } \over 4}$$

Solution

Beats period$$=\cfrac{1}{30-20}=0.1second\\ \Delta Y=\cfrac{2\pi}{T}\Delta t\\ \quad=\cfrac{2\pi}{0.1}\times0.6=2\pi\times6=12\pi\\ \therefore\Delta Y=12\pi$$
Question 4
1 / -0

The displacements of two particles executing $$SHM$$ on the same line are given as $$y_{1}=a\sin \left (\dfrac {\pi}{2}t+ \phi \right)$$ and $$y_{2}=b\sin \left (\dfrac {2\pi}{3}t+ \phi \right)$$. The phase difference between them at $$t=1\ s$$ is:

A
$$\pi$$

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

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

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

Solution

At $$t=1 sec, y_1=a\sin(\pi /2+\phi)$$.....(1)
$$y_2=b\sin(2\pi/3+\phi)=b\sin(\dfrac{\pi}{2}+\dfrac{\pi}{6}+\phi)...$$(2)
From (1) and (2) it is clear that,
Phase difference$$=\dfrac{\pi}{6}$$
Question 5
1 / -0

The displacement y of a particle varies with time t as $$y = 4 \sin\omega t\ \cos \omega t $$ cm where t is in seconds. The motion of the particle is

A
simple harmonic with amplitude 2 cm

B
simple harmonic with period $$\cfrac{\pi }{\omega } $$

C
simple harmonic with period $$\cfrac{2\pi }{\omega } $$

D
both a & b

Solution

$$y=4\sin\omega t\cos\omega t$$
$$=2\sin 2\omega t$$
$$A=2cm,$$ Performing SHM.
Question 6
1 / -0

Period of a particle performing S.H.M is 4 s. It starts motion from mean position. After 2/3 s, its velocity is 6.28 cm/s. The amplitude of S.H.M is:

A
(a) 8 cm

B
(b) $$\frac{4}{\sqrt{3}}$$

C
(c) $$\frac{8}{\sqrt{3}}$$

D
(d) 4 cm

Solution
Question 7
1 / -0

A point source emits sound equally in all directions in a non-absorbing medium. Two points $$P$$ and $$Q$$ are at distance of $$2\ m$$ and $$3\ m$$ respectively from the source. The ratio of the intensities of the wave at $$P$$ and $$Q$$ is:

A
$$9:4$$

B
$$2:3$$

C
$$3:2$$

D
$$4:9$$

Solution

intensity is inversely proportional to $$r^2$$
$$I_1:I_2=r_2^2:r_1^2=3^2:2^2=9:4$$
Question 8
1 / -0

Two coherent transverse waves of amplitude 2A and A of same frequency propagated respectively along $$+ve$$ and $$-ve$$ x-axis are superimposed. The amplitude of the resultant wave at a distance of 5 cm from the origin,$$(given K = \cfrac{\pi}{4}cm^{-1})$$ is

A
$$\sqrt{5A}$$

B
4 A

C
5 A

D
$$\sqrt{3A}$$
Question 9
1 / -0

Two particles $$P$$ and $$Q$$ describe SHM of same amplitude $$a$$ and frequency $$f$$ along the same straight line. The maximum distance between two particles is $$a\sqrt{2}$$. The initial phase difference between the particles is:

A
$$\text {Zero}$$

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

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

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

Solution

let the phase difference = z
the equation of displacement of the particle p from origin
$$y=rsin(\omega t + z)$$
the equation of displacement of the particle q from origin
$$x=rn(\omega t)$$
$$2rsin(\dfrac{z}{2})= \dfrac{r}{\sqrt{2}}$$
given max distance between the particles is $$\sqrt{2}r$$
hence maximum phase difference is zero
Question 10
1 / -0

Two waves of amplitudes $$A_0$$ and $$xA_0$$pass through a region.If $$x>1$$,what is the difference in the maximum and minimum resultant amplitude is :

A
$$(x+1)A_0$$

B
$$(x-1)A_0$$

C
$$2xA_0$$

D
$$2A_0$$

Solution

Given, Waves having the amplitude $$A_0 \ and xA_0$$
So, the maximum amplitude $$=A_0+xA_0 $$ (it is the constructive superposition),
The minimum amplitude $$=xA_0-A_0 $$ (it is a destructive superposition).
Therefore the difference between maximum and minimum amplitude $$=xA_0+A_0-(xA_0-A_0)=2A_0$$

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Waves Test - 49

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Waves Test - 49

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SHARING IS CARING

Question 1

$$5\sqrt { 2+\sqrt { 3 } } mm$$

$$5\sqrt { 3 } mm$$

$$5mm$$

$$5\sqrt { 2 }mm $$

Solution

Question 2

5 cm

6 cm

7 cm

9 cm

Solution

Question 3

$$12\pi $$

$${\pi \over 2}$$

$$\pi $$

$${{3\pi } \over 4}$$

Solution

Question 4

$$\pi$$

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

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

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

Solution

Question 5

simple harmonic with amplitude 2 cm

simple harmonic with period $$\cfrac{\pi }{\omega } $$

simple harmonic with period $$\cfrac{2\pi }{\omega } $$

both a & b

Solution

Question 6

(a) 8 cm

(b) $$\frac{4}{\sqrt{3}}$$

(c) $$\frac{8}{\sqrt{3}}$$

(d) 4 cm

Solution

Question 7

$$9:4$$

$$2:3$$

$$3:2$$

$$4:9$$

Solution

Question 8

$$\sqrt{5A}$$

4 A

5 A

$$\sqrt{3A}$$

Question 9

$$\text {Zero}$$

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

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

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

Solution

Question 10

$$(x+1)A_0$$

$$(x-1)A_0$$

$$2xA_0$$

$$2A_0$$

Solution

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