Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, c) y (x, t)=30sin (36t+0.018x+ $\frac{\pi }{4}$ )

Y=asin (wt+kx+ $\varnothing$ )

(a) as the equation involves positive sign with x . hence the wave is travelling from right to left. Hence option a is correct.

(b) W=36

2 $\pi v=36$

$v=\frac{36}{2\pi }$ = $\frac{18}{\pi }$

$k=0.018=\frac{2\pi }{\lambda }$

$\frac{2\pi v}{v\lambda }=0.018$

w/v=0.018

v=2000cm/s=20m/s

(c) 2 $\pi v=36$

$v=\frac{18}{\pi }$ = 5.7Hz

(d) $\frac{2\pi }{\lambda }=0.018$ , $\lambda =2\pi$ /0.0018cm=3.48cm

This is a multiple choice answer as classified in NCERT Exemplar

(c) Let frequency of the source is n_o

Let the speed of sound wave in the medium is v

As observer is stationary

Apparent frequency n_a= $\frac{v}{v-v_s}{n}_o$

$\frac{v}{v-v_s}{n}_o=n_a{n}_o$

When the train is going away from the observer

Apparent frequency n_a= $\frac{v}{v-v_s}{n}_o$ = $n_a<n_o$

This is a multiple choice answer as classified in NCERT Exemplar

(b) m=2.5kg

$\mu =$ mass per unit length

=m/l =2.5/20=125/10=0.125kg/m

V= $\sqrt[]{\frac{T}{\mu }}=\sqrt[]{\frac{200}{0.125}}$

L=v $*t$

20= $\sqrt[]{\frac{125}{2*{10}^5}}=20*\sqrt[]{\frac{25*5}{2*{10}^5}}$

= 20 $*\sqrt[]{\frac{25}{0.4*{10}^5}}=\frac{1}{2}$

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(b) Amplitude of reflected wave

A_f= $\frac{2}{3}*A_i$ =2 (0.6)/3=0.4 units

Equation of incident wave

Y_i=0.6sin2 $\pi \left(t-\frac{x}{2}\right)$

For reflected wave 

y_r=A_rsin2 $\pi$ (t+x/2+ $\pi$ )

Y_r=-0.4sin2 $\pi \left(t+\frac{x}{2}\right)$

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(d) Due to compression and rarefactions density of the medium changes.at compressed regions density is maximum and at rarefactions density is minimum

As density is changing so boyle, s law not obeyed

Bulk modulus remains same

The time of compression and rarefaction is too small so no transfer of heat.

This is a multiple choice answer as classified in NCERT Exemplar

(c) When mechanical transverse wave propagates through a medium, the constituents of the medium oscillate perpendicular to wave motion causing change in shape. That is each element of the medium is subjected to shearing stress. Solids and strings have shear, that is why tey bear stress. But in fluids there is no fixed shape so transverse waves does not formed in fluid.

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(b) Propagation of longitudinal waves through a medium leads to transmission of energy through the medium without matter being transmitted. So no movement of energy and matter hence momentum

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(c) Speed of sound wave in a medium v $\propto \sqrt[]{T}$

Clearly when temperature changes speed also changes

As v= $v\lambda$ as frequency remains constant and speed changes so wavelength changes

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(c) Due to presence of moisture density of air decreases.

As we know speed of sound in air v= $\sqrt[]{\frac{\gamma P}{\rho }}$

So v $\propto \frac{1}{\sqrt[]{\rho }}$

v_2/v₁= $\sqrt[]{\frac{{\rho }_2}{{\rho }_1}}$

_{$\rho$ 2} (moist air)< $\rho$ ₁ (dry air)

So v₂>v₁

So speed of sound wave increases with increase in humidity

This is a multiple choice answer as classified in NCERT Exemplar

(c) Let frequency of the two will be  $vandv\text{'}$  and it is same in both medium

So v/ $\lambda$ = v'/ $\lambda$

$\lambda \text{'}$ = v' $\lambda$ /v

$\lambda \text{'}$ = 2v $\lambda$ /v=2 $\lambda$

Where $\lambda$ and $\lambda$ ' and v and v' are wavelength and speeds of first and second medium respectively