Waves

d₁ = 2 m, d₂ = 3 m

$Intensity \propto {1 \over {{{\left( {distance} \right)}^2}}}$

${I_1} \propto {1 \over {{2^2}}}$ and ${I_2} \propto {1 \over {{3^2}}}$

$ \therefore {{{I_1}} \over {{I_2}}} = {9 \over 4}$

Let f' be the frequency of sound heard by cliff.

$ \therefore $ $f = {{vf} \over {v - {v_c}}}$   ...(1)

Now, for the reflected wave, cliff acts as a source,

$ \therefore $ $2f = {{f'\left( {v + {v_c}} \right)} \over v}$   ...(2)

$2f = {{f\left( {v + {v_c}} \right)} \over {v - {v_c}}}$

$ \Rightarrow 2v - 2{v_c} = v + {v_c}$

$ \Rightarrow {v \over 3} = {v_c}$

y₁ = 10^–6sin[100t + (x/50) + 0.5]

y₂ = 10^–6cos[100t + (x/50)]
[using cosx = sin(x + $\pi $/2)]

= 10^–6sin[100t + (x/50) + $\pi $/2]

= 10^–6sin[100t + (x/50) + 1.57]

The phase difference = 1.57 – 0.5 = 1.07

[or using sinx = cos($\pi $/2 – x). We get the same result.]

Apparent frequency, $f' = {{v + {v_0}} \over v}f$

$ = {{v + (1/5)v} \over v}f = 1.2f$

Wavelength does not change by motion of observer.

The whistle is revolving in a circle of radius 50 cm. So the source (whistle) is moving and the observer is fixed.


The minimum frequency will be heard by the observer when the linear velocity of the whistle (source) will be in a direction as shown in the figure, i.e. when the source is receding.

The apparent frequency heard by the observer is then given by $\upsilon ' = \upsilon \left( {{v \over {V + v}}} \right)$

where V and v are the velocities of sound and source respectively and $\upsilon $ is the actual frequency.

Now $v = r\omega = 0.5 \times 20 = 10\,m/s$

V = 340 m/s, $\upsilon $ = 385 Hz.

$ \therefore $ $\upsilon ' = 385 \times {{340} \over {340 + 10}} = 374$ Hz

The equation of progressive wave travelling in positive x-direction is given by

$y = a\sin {{2\pi } \over \lambda }\left( {vt - x} \right)$

Here a = 0.2 m, v = 360 m/sec, $\lambda $ = 60 m

$ \therefore y = 0.2\sin {{2\pi } \over {60}}\left( {360t - x} \right)$

$ = 0.2sin\left[ {2\pi \left( {6t - {x \over {60}}} \right)} \right]$

Comparing the given equation with general equation,
$y = a\sin 2\pi \left( {{t \over T} - {x \over \lambda }} \right)$, we get

$T = {{2\pi } \over {100}}$ and $\lambda = 20\pi $

$ \therefore v = \upsilon \lambda = {{100} \over {2\pi }} \times 20\pi = 1000\,m/s$

Resultant amplitude = 2a (1 + cos$\phi $) = a

$ \therefore $ $\left( {1 + \cos \phi } \right) = 1/2;$

$cos\phi = - {1 \over 2};\phi = {{2\pi } \over 3}$

$n = {1 \over {2l}}\sqrt {{T \over {\pi {r^2}\rho }}} $

$\rho _1^{'} = {\rho \over 2};T' = 2T$

D' = 2D or r' = 2r

$n' = {1 \over {2l}}\sqrt {{{2T} \over {\pi {{\left( {2r} \right)}^2}{\rho \over 2}}}} $

After solving, $n' = {1 \over {2l}}\sqrt {{{2T} \over {\pi {r^2}l}}} = n$

As $n \propto \left( {1/l} \right)$

$l = {l_1} + {l_2} + {l_3}$

$ \therefore {1 \over n} = {1 \over {{n_1}}} + {1 \over {{n_2}}} + {1 \over {{n_3}}}$

Given : x = acos($\omega $t + $\delta $)
and y = acos($\omega $t + $\alpha $)     ...(i)

where, $\delta = \alpha + \pi /2$

$ \therefore {\rm{x = acos(}}\omega {\rm{t + }}\alpha {\rm{ + }}\pi {\rm{/2)}}$

= –asin($\omega $t + $\alpha $) ...(ii)

Given the two waves are acting in perpendicular direction with the same frequency and phase difference $\pi $/2.

From equations (i) and (ii),
x² + y² = a²
which represents the equation of a circle.

$f' = {{v - u} \over v}f;f'' = {{v + u} \over v}f$

Number of beats = $f'' - f' = {{2u} \over \lambda }$