Waves Test

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Waves Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

Two sound waves are represented by the following equations
$$\quad y_1 = 10\sin(3\pi - 0.03x)$$ and
$$\quad y_2 = 5\sin(3\pi t - 0.03x) + 5\sqrt3\cos(3\pi t - 0.03 x)$$
Then, the ratio of their amplitudes is given by

A
$$1:1$$

B
$$1:2$$

C
$$2:1$$

D
$$2:5$$

Solution

mistake in the question .
if it is $$ 5 \sqrt 3$$ , then the below equation is a combination of two waves with pase difference of $$ \frac {\pi} {2}. $$
hence its amplitude is $$ \sqrt {{5}^{2} + {(5\sqrt3)}^{2}}$$ = 10 . hence ratio is 1:1
Question 2
1 / -0

If the frequency of a sound wave is doubled then the velocity of sound will be

A
zero

B
half

C
double

D
unchanged

Solution

If the velocity of sound wave were frequency dependent, then the sound coming from different instruments of orchestra will arrive to our ears at different time and will not listen proper music. Speed of sound waves are frequency independent. option "D" is correct.
Question 3
1 / -0

A boat at anchor is rocked by a wave whose crests are $$100\space m$$ apart and whose velocity is $$25\space ms^{-1}$$.how often does the crest reach the boat?

A
$$2500\space s$$

B
$$1500\space s$$

C
$$4.0\space s$$

D
$$0.25\space s$$

Solution

Distance between consecutive crests is the wavelength of the wave.
Here given is $$\lambda=100m$$
$$v=\lambda\nu$$
$$\implies \nu=\dfrac{v}{\lambda}$$
$$\implies \nu=\dfrac{100m}{25m/s}=4s$$
Question 4
1 / -0

Two waves of same amplitude and same frequency reach a point in a medium simultaneously. The phase difference between them for resultant amplitude to be zero, will be

A
$$4\pi$$

B
$$2\pi$$

C
$$\pi$$

D
$$0^{\small\circ}$$

Solution

The equation of the two superimposing waves can be give by
$$y_1=a\sin(\omega t)$$ and $$y_2=a\sin(\omega t +\phi)$$. So, the equation of the resultant wave is $$y=y_1+y_2=a(\sin(\omega t)+\sin(\omega t+\phi))=0\Rightarrow \sin(\omega t)=-\sin(\omega t+\phi)\Rightarrow \phi=\pi, 3\pi...$$.
Question 5
1 / -0

The resultant amplitude, when two waves of same frequency but with amplitudes $$a_1$$ and $$a_2$$ superimpose with a phase difference of $$\pi/2$$ will be

A
$$a_1^2 + a_2^2$$

B
$$\sqrt{a_1^2 + a_2^2}$$

C
$$a_1 - a_2$$

D
$$a_1 + a_2$$

Solution

The equation of the two superimposing waves can be give by
$$y_1=a_1\sin(\omega t)$$ and $$y_2=a_2\sin(\omega t +\pi/2)=a_2\cos(\omega t)$$.
So, the equation of the resultant wave is $$y=y_1+y_2=a_1\sin(\omega t)+a_2\cos(\omega t)$$.
Assuming $$a_1=a\cos(\phi)$$ and $$a_2=a\sin(\phi)$$, we get
$$y=a\sin(\omega t+\phi)$$.
Here $$a^2_1+a^2_2=a^2(\sin^2(\phi)+\cos^2(\phi))=a^2\Rightarrow a=\sqrt{a^2_1+a^2_2}$$.
Question 6
1 / -0

A rod $$70\space cm$$ long is clamped from middle. The velocity of sound in the material of the rod is $$3500\space ms^{-1}$$. The frequency of fundamental note produced by it is :

A
$$3500\space Hz$$

B
$$2500\space Hz$$

C
$$1250\space Hz$$

D
$$700\space Hz$$

Solution

SInce rod is clamped at middle, therefore only nodes can form at that point, and at free end only antinode can form, therefore for fundamental mode of frequency,

$$\lambda/4=L/2$$

$$\lambda=2L=2\times 70 cm=1.4 m$$

$$v=3500\ ms^{-1}$$ is given,

We know, $$v=f \times \lambda$$ or $$f = \dfrac{v}{\lambda}=3500/1.4=2500 Hz$$

Option "B" is correct.
Question 7
1 / -0

$$P$$ is the junction of two wires $$A$$ and $$B$$. $$B$$ is made of steel and is thicker while $$A$$ is made of aluminium and is thinner as shown. If a wave pulse as shown in the figure approaches $$P$$, the reflected and transmitted waves from $$P$$ are respectively :

A

B

C

D

Solution

When a wave incidents on a denser medium from a rarer medium, it goes through $$\pi$$ phase change for the reflected beam and $$0$$ phase change for the transmitted beam.
Question 8
1 / -0

For a pulse moving in a heavy string the junctions of the string behaves as a

A
perfectly rigid end

B
free end

C
partially rigid end

D
rigid end

Solution

Waves in a heavy string move slower compared to thinner strings. And the junction acts as a perfectly rigid end because of the change in the density and also the lesser speed of the wave.
Question 9
1 / -0

A wave represented by $$y = 100\sin(ax+bt)$$ is reflected from a dense plane at the origin. If $$36\mbox{%}$$ of energy is lost and rest of the energy is reflected then the equation of the reflected wave will be:

A
$$y = -80\sin(ax+bt)$$

B
$$y = -8.1\sin(ax+bt)$$

C
$$y = -10\sin(ax+bt)$$

D
$$y = -8.2\sin(ax+bt)$$

Solution

Remaining energy of the reflected beam is $$(100-36)\%=64\%$$.
So intensity of the reflected beam can be given by,
$$I'=64\% \text{ of } I_0$$
or, $$I'=\dfrac{64}{100}A^2=\dfrac{64}{100}(100)^2=6400$$, where A is the amplitude of the incident wave.
Amplitude of the reflected beam A' can now be derived as
$$I'=A'^2\Rightarrow A' =\sqrt{I'}=80$$.
Considering the change in phase due to reflection from a denser medium, we get the equation of the reflected beam as :
$$y=80\sin(ax+bt+\pi)=-80\sin(ax+bt)$$.
Question 10
1 / -0

If the wavelength of a wave is decreased by $$20\mbox{%}$$ then its frequency will become :

A
$$20\mbox{%}\space less$$

B
$$25\mbox{%}\space more$$

C
$$20\mbox{%}\space more$$

D
$$25\mbox{%}\space less$$

Solution

Velocity of sound in the air does not change with frequency it means that the product of wavelength and frequency remains constant for different sounds.

$$v=f_1 \times \lambda_1 = f_2 \times \lambda_2 $$

if, $$\lambda_2= 0.8 \lambda_1$$,

$$f_2 = f_1 \dfrac{\lambda_1}{ \lambda_2}= f_1 (1/0.8)=1.25 \lambda_1$$

option "B" is correct.