Question

The equation of a travelling sound wave is $y = 6.0\sin (600t - 1.8x)$ where $y$ is measured in ${10^{ - 5\,}}\,m$, $t$ in second and $x$ in meter.
(a) Find the ratio of the displacement amplitude of the particles to the wavelength of the wave.
(b) Find the ratio of the velocity amplitude of the particles to the wave speed.

Answer 1

The equation for the oscillating wave is given as $y = A\sin (\omega t - K) + b$. where. $A$ = Displacement amplitude, $\omega$ is the wavelength, $t$ is the time, $K$ is some arbitrary constant representing time offset, $b$ is arbitrary displacement offset. Comparing the given equation with this equation, we can easily find the given values. Ratio of displacement is given as $\dfrac{A}{\lambda }$. The maximum velocity of a particle is known as velocity amplitude. The wave speed is calculated using the relation between velocity, frequency and wavelength.

Complete step by step solution:
The given relation is: $y = 6.0\sin (600t - 1.8x)$ comparing this with the equation of oscillating wave, $x = A\sin (\omega t - K) + b$ we have:
$A = 6.0$
$\omega = 600$
$K = 1.8$
Since, we are given that the displacement is in the order of ${10^{ - 5\,}}\,m$. Therefore, we can have the amplitude as:
$A = 6.0 \times {10^{ - 5}}\,m$
The wavelength $\lambda$ is given as:
$\lambda = \dfrac{{2\pi }}{K}$
Substituting the values, we get:
$\lambda = \dfrac{{2\pi }}{{1.8}}$ …………………………………...equation $(1)$
The displacement amplitude is given as $\dfrac{A}{\lambda }$ , substituting the value of amplitude and wavelength, we get:
$\dfrac{A}{\lambda } = \dfrac{{6 \times {{10}^{ - 5}}}}{{\dfrac{{2\pi }}{{1.8}}}}$
$\Rightarrow \dfrac{A}{\lambda } = \dfrac{{5.4 \times {{10}^{ - 5}}}}{\pi }$
$\Rightarrow \dfrac{A}{\lambda } = 1.7 \times {10^{ - 5}}$
Therefore, the ratio of the displacement amplitude of the particles to the wavelength of the wave is $1.7 \times {10^{ - 5}}$
To find the ratio of the velocity amplitude of the particles to the wave speed, let us find the velocity of the particle and the wave speed.
As, velocity is rate of change of displacement thus, differentiating the equation of displacement we will get the velocity of the particle.
$v = \dfrac{{dy}}{{dt}}$
Here, $v$ is the velocity of the particle.

$\Rightarrow v = \dfrac{{d\left( {6.0\sin (600t - 1.8x)} \right)}}{{dt}}$
$\Rightarrow v = 3600\cos (600t - 1.8x) \times {10^{ - 5}}$
This velocity will be maximum when the value of cosine is maximum. The maximum value of cosine is $1$ . Therefore, the maximum velocity will be
$\Rightarrow v = 3600(1) \times {10^{ - 5}}$
$\Rightarrow v = 3600 \times {10^{ - 5}}\,m\,{s^{ - 1}}$ ………………………….equation $(2)$
Now, for the speed of the wave, we have
$\omega = 600$
But $\omega = 2\pi f$ , $f$ is the frequency. Thus, the frequency will be:
$f = \dfrac{\omega }{{2\pi }}$
Substituting the values, we get
$f = \dfrac{{600}}{{2\pi }}$
The wave speed ${v_s}$ is given ${v_s} = f \times \lambda$ .
$\Rightarrow {v_s} = \dfrac{{600}}{{2\pi }} \times \dfrac{{2\pi }}{{1.8}}$
$\Rightarrow {v_s} = \dfrac{{1000}}{3}\,m\,{s^{ - 1}}$ …………………..equation $(3)$
Dividing equation $2$ by equation $3$ , we will get the ratio of the velocity amplitude of the particles to the wave speed.
$\dfrac{v}{{{v_s}}} = \dfrac{{3600 \times {{10}^{ - 5}}}}{{\dfrac{{1000}}{3}}}$
$\dfrac{v}{{{v_s}}} = 1.08 \times {10^{ - 4}}$

Therefore, the ratio of the displacement amplitude of the particles to the wavelength of the wave is $\dfrac{A}{\lambda } = 1.7 \times {10^{ - 5}}$ and ratio of the velocity amplitude of the particles to the wave speed is $\dfrac{v}{{{v_s}}} = 1.08 \times {10^{ - 4}}$.

Note: The ratio will be a dimensionless quantity. Comparing the term with the general equation, we get the value of various variables. The equation of the velocity of the particle is obtained by differentiating the equation of the displacement with respect to time. The wave speed is calculated using the relation between speed, frequency and wavelength. The magnitude of displacement is given as ${10^{ - 5\,}}\,m$.