Waves

y = 0.25sin(10$\pi $x – 2$\pi $t)

${y_{\max }} = 0.25$

$k = {{2\pi } \over \lambda } = 10\pi \Rightarrow \lambda = 0.2\,m$

$\omega = 2\pi f = 2\pi \Rightarrow f = 1\,Hz$

The sign is negative inside the bracket. Therefore this wave travels in the positive x-direction.

Light waves are electromagnetic waves. Light waves are transverse in nature and do not require a medium to travel, hence they can travel in vacuum. Sound waves are longitudinal waves and require a medium to travel. They do not travel in vacuum.

$y\left( {x,t} \right) = 8.0\sin \left( {0.5\pi x - 4\pi t - {\pi \over 4}} \right)$

Compare with a standard wave equation,

$y = a\sin \left( {{{2\pi x} \over \lambda } - {{2\pi t} \over T} + \phi } \right)$

we get ${{2\pi } \over \lambda } = 0.5\pi \Rightarrow \lambda = {{2\pi } \over {0.5\pi }} = 4m$

${{2\pi } \over T} = 4\pi \Rightarrow T = {{2\pi } \over {4\pi }} = {1 \over 2}\sec $

$\upsilon = 1/T = 2\,Hz$.

Wave velocity, $v = \lambda \upsilon = 4 \times 2 = 8$ m/sec.

Reverberation time is defined as the time during which the intensity of sound in the auditorium becomes one millionth of the initial intensity.

Sabine has shown that standard reverberation time for an auditorium is given by the formula

${T_R} = K{V \over {\alpha S}}$

Here, V is volume of the auditorium, S is the surface area.

$ \therefore {T_R} = {{K.V} \over {\alpha S}} = 1$ (given)

$ \Rightarrow {T_R} = {{K.{l^3}} \over {6\alpha {l^2}}} = {K \over {6\alpha }}l$ ${T_R} = {{K.{l^3}} \over {6\alpha {l^2}}} = {K \over {6\alpha }}l$
(Assuming auditorium to be cubic in shape)

$ \therefore {T_R} \propto l$

If dimension is doubled, reverberation time t will be doubled.
So, New T_R = 2 sec.

Frequencies of sound waves are
${{330} \over 5}$ & ${{330} \over {5.5}}$

i.e., 66 Hz and 60 Hz

Frequencies of beat = 66 – 60 = 6 per second

Y₁ = 4 sin500$\pi $t, Y₂ = 2 sin506$\pi $t

${\omega _1} = 500\pi $, ${\omega _1} = 2\pi \upsilon $,
${\upsilon _1} = 250,{\upsilon _2} = 253$

$\upsilon = {\upsilon _2} - {\upsilon _1} = 253 - 250 = 3$ beats/s

$ \therefore $ Number of beats per minute = 3 × 60 = 180

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