Waves Test

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Waves Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

An SHM is represented by $$x=5\sqrt{2}(sin\, 2\pi t+cos\, 2\pi t)$$. The amplitude of the SHM is:

A
10 cm

B
20 cm

C
$$5\sqrt{2}$$ cm

D
50 cm

Solution

Here, $$x = 5\sqrt{2}(sin\, 2\pi t+cos\,2\pi t)$$
$$\Rightarrow x=5\sqrt{2} sin \,2\pi t+5\sqrt{2} cos\, 2\pi t$$ ...(i)
The standard equation of simple harmonic motion is given by
$$x=A_1sin\,\omega t+A_2cos\,\omega t$$ ...(ii)
Now, comparing Eqs. (i) and (ii), we obtain
$$A_1=\sqrt{A_1^2+A_2^2}=\sqrt{(5\sqrt{2})^2+(5\sqrt{2})^2}$$
$$=\sqrt{50+50}$$
$$=10 cm$$
Question 2
1 / -0

A pulse of a wave train travels along a stretched string and reaches the fixed end of the string. It will be reflected back with :

A
a phase change of $${180}^{o}$$ with velocity reversed

B
the same phase as the incident pulse with no reversal of velocity

C
a phase change of $${180}^{o}$$ with no reversal of velocity

D
the same phase as the incident pulse but with velocity reversed

Solution

A pulse of wave train when travels along a stretched string and reaches the fixed end of the string, then it will be reflected back to the same medium and the reflected ray suffers a phase change of $$\pi$$ with the incident wave but wave velocity after reflection reverses its direction.
Question 3
1 / -0

The amplitude of two waves are in ratio 5 : 2. If all other conditions for the two waves are same, then what is the ratio of their energy densities?

A
5 : 2

B
5 : 4

C
4 : 5

D
25 : 4

Solution

Energy density of wave is given by
$$u=2\pi^2n^2pA^2$$
or $$u \propto A^2$$ (As n and p are constant)
$$\therefore \dfrac{u_1}{u_2}=\dfrac{A_1^2}{A_2^2}=\dfrac{5^2}{2^2}$$
So, $$u_1:u_2=25:4$$
Question 4
1 / -0

A progressive wave of frequency $$500\ Hz$$ is travelling with a speed of $$330\ m/s$$ in air. The distance between the two points which have a phase difference of $$\displaystyle { 30 }^{ \circ }$$ is:

A
$$0.11\ m$$

B
$$0.055 \ m$$

C
$$0.22\ m$$

D
$$0.025\ m$$

Solution

Wavelength of the wave is $$\lambda =\dfrac { v }{ f } =\dfrac { 330 }{ 500 }\ m =\dfrac { 33 }{ 50 }\ m $$

$$k=\dfrac { 2\pi }{ \lambda } =\dfrac { 100\pi }{ 33 } $$

Since, phase difference is given by $$k\Delta x$$

$$\Delta \phi=k\Delta x=\dfrac { 100\pi }{ 33 } \Delta x=\dfrac { \pi }{ 6 } $$

$$\Delta x=0.055\ m$$
Question 5
1 / -0

The displacement of a particle varies according to the relation $$x=4(\cos {\pi t}+\sin {\pi t})$$. The amplitude of the particle is:

A
$$-4$$

B
$$4$$

C
$$4\sqrt {2}$$

D
$$8$$

Solution

Hint: Here, we have to apply trigonometric formulas

Solution:

Step1: Simplify the given displacement

$$x = 4\left( {cos\pi t{\text{ }} + {\text{ }}sin\pi t} \right).....(1)$$

Multiply & divide each term of R.H.S. of equation (1) by $$\sqrt 2 $$ we get,

$$ \Rightarrow \dfrac{4}{{\sqrt 2 }} \times \sqrt 2 \left( {cos\pi t{\text{ }} + sin\pi t} \right)$$

$$ \Rightarrow 4\sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }}\cos \pi t + \dfrac{1}{{\sqrt 2 }}\sin \pi t} \right).....(2)$$

Step2: Apply required trigonometric formula
we know that, $$\dfrac{1}{{\sqrt 2 }} = \sin \dfrac{\pi }{4}$$ and $$\dfrac{1}{{\sqrt 2 }} = \cos \dfrac{\pi }{4}$$

Now, equation (2) becomes,

$$ \Rightarrow 4\sqrt 2 \left( {\cos \pi t\sin \dfrac{\pi }{4} + \sin \pi t\cos \dfrac{\pi }{4}} \right).....(3)$$

Also, $$\sin (a + b) = \sin a\cos b + \cos a\sin b$$

Now, equation (3) becomes,

$$x = 4\sqrt 2 \sin \left( {\pi t{\text{ }} + \dfrac{\pi }{4}} \right).....(4)$$

Step3: Find Amplitude ‘A’
Comparing above equation (4) with,

$$x = A(\sin \omega t + \Phi )$$

We get, Amplitude, $$A = 4\sqrt 2$$

Hence, option (C) is correct.
Question 6
1 / -0

If two waves of same frequency and amplitude respectively on superposition produce a resultant disturbance of the same amplitude, the wave differ in phase by :

A
$$\pi $$

B
zero

C
$$\pi /3$$

D
$$2\pi /3 $$

Solution

Resultant amplitude due to superposition of two waves with phase difference $$\phi$$ is given by
$$A^2=A_1^2+A_2^2+2A_1A_2cos\phi$$
It is given that $$A_1=A_2=A$$
Thus $$A^2=A^2+A^2+2A^2cos\phi$$
$$\implies cos\phi=-\dfrac{1}{2}$$
$$\implies \phi=\dfrac{2\pi}{3}$$
Question 7
1 / -0

Two progressive waves having equation $$x_{1} = 3\sin \omega \tau$$ and $$x_{2} = 4\sin (\omega \tau = 90^{\circ})$$ are super imposed. The amplitude of the resultant wave is

A
$$5\ unit$$

B
$$1\ unit$$

C
$$3\ unit$$

D
$$4\ unit$$

Solution

$$x_{1} = 2\sin \omega t$$
$$x_{2} = 4\sin (\omega t + 90^{\circ})$$
The phase difference between the two waves is $$90^{\circ}$$.
So, resultant amplitude
$$a = \sqrt {(3)^{2} + (4)^{2}}$$
$$= \sqrt {9 + 16} = \sqrt {25}$$
$$= 5\ unit$$
Question 8
1 / -0

In a medium in which, a transverse progressive wave travelling, the phase difference between the points with a separation of $$1.25\ cm$$ is $$\displaystyle { \pi }/{ 4 }$$. If the frequency of wave $$1000\ Hz$$, the velocity in the medium is :

A
$$\displaystyle { 10 }^{ 4 }\ { ms }^{ -1 }$$

B
$$\displaystyle 125\ { ms }^{ -1 }$$

C
$$\displaystyle 100\ { ms }^{ -1 }$$

D
$$\displaystyle 10\ { ms }^{ -1 }$$

Solution

Let $$x_1$$ and $$x_2$$ be two points,
then, the phase difference between two points $$x_1 and \ x_2$$ is $$\dfrac{2\pi}{\lambda}(x_2-x_1)$$
$$\implies \dfrac{2\pi}{\lambda}(1.25\times 10^{-2})=\dfrac{\pi}{4}$$
$$\implies \lambda=0.1m$$
Thus, speed of wave in the medium =$$\lambda\nu=0.1\times 1000=100\ m/s$$
Question 9
1 / -0

Five sinusoidal waves have the same frequency $$500Hz$$ but their amplitudes are in the ratio $$2:\cfrac{1}{2}:\cfrac{1}{2}:1:1$$ and their phase angles $$0,\cfrac{\pi}{6},\cfrac{\pi}{3},\cfrac{\pi}{2}$$ and $$\pi$$ respectively. The phase angle of resultant wave obtained by the superposition of these five waves is:

A
$${30}^{o}$$

B
$${45}^{o}$$

C
$${60}^{o}$$

D
$${90}^{o}$$

Solution

The phasors of five waves can be represented as
$$x_1=2\hat i$$
$$x_2=\sqrt 3/4 \hat i + 1/4 \hat j$$
$$x_3=1/4 \hat i + \sqrt 3 /4 \hat j$$
$$x_4=\hat j$$
$$x_5=-\hat i$$

Resultant $$x=\sum_{i=1}^5 x_i$$
$$x=\cfrac{5+\sqrt 3}{4}\hat i + \cfrac{5+\sqrt 3}{4}\hat j$$
Phase angle of resultant, $$\theta =tan^{-1}\dfrac{(5+\sqrt 3)/4}{(5+\sqrt 3)/4}$$
$$\Rightarrow \theta = 45^o$$
Question 10
1 / -0

If n represents the order of a half period zone the area of this zone is approximately proportional to $$n^{m}$$ where m is equal to

A
zero

B
half

C
one

D
two

Solution

Area of half period zone is independent of order of zone. Therefore, m is equal to zero in $$n^m$$.

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Re-Attempt Weekly Quiz Competition

Waves Test - 37

Result

Waves Test - 37

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SHARING IS CARING

Question 1

10 cm

20 cm

$$5\sqrt{2}$$ cm

50 cm

Solution

Question 2

a phase change of $${180}^{o}$$ with velocity reversed

the same phase as the incident pulse with no reversal of velocity

a phase change of $${180}^{o}$$ with no reversal of velocity

the same phase as the incident pulse but with velocity reversed

Solution

Question 3

5 : 2

5 : 4

4 : 5

25 : 4

Solution

Question 4

$$0.11\ m$$

$$0.055 \ m$$

$$0.22\ m$$

$$0.025\ m$$

Solution

Question 5

$$-4$$

$$4$$

$$4\sqrt {2}$$

$$8$$

Solution

Question 6

$$\pi $$

zero

$$\pi /3$$

$$2\pi /3 $$

Solution

Question 7

$$5\ unit$$

$$1\ unit$$

$$3\ unit$$

$$4\ unit$$

Solution

Question 8

$$\displaystyle { 10 }^{ 4 }\ { ms }^{ -1 }$$

$$\displaystyle 125\ { ms }^{ -1 }$$

$$\displaystyle 100\ { ms }^{ -1 }$$

$$\displaystyle 10\ { ms }^{ -1 }$$

Solution

Question 9

$${30}^{o}$$

$${45}^{o}$$

$${60}^{o}$$

$${90}^{o}$$

Solution

Question 10

zero

half

one

two

Solution

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