Question

A sinusoidal wave travelling in the positive direction on stretched string has amplitude $20\,{\text{cm}}$, wavelength ${\text{1}}\,{\text{m}}$ and wave velocity ${\text{5 m}}{{\text{s}}^{ - 1}}$. At $x = 0$ and $t = 0$, it is given that $y = 0$ and $\dfrac{{dy}}{{dt}} < 0$. Find the wave function $y(x,t)$
A. $y(x,t) = \left( {0.2{\text{m}}} \right){\text{sin}}\left[ {\left( {{\text{2}}\pi {{\text{m}}^{ - 1}}} \right)x + \left( {10\pi {{\text{s}}^{ - 1}}} \right)t} \right]{\text{m}}$
B. $y(x,t) = \left( {0.2{\text{m}}} \right)\cos \left[ {\left( {10\pi {{\text{s}}^{ - 1}}} \right)t + \left( {2\pi {{\text{m}}^{ - 1}}} \right)x} \right]{\text{m}}$
C. $y(x,t) = \left( {0.2{\text{m}}} \right)\sin \left[ {\left( {2\pi {{\text{m}}^{ - 1}}} \right)x - \left( {10\pi {{\text{s}}^{ - 1}}} \right)t} \right]{\text{m}}$
D. $y(x,t) = \left( {0.2{\text{m}}} \right)\sin \left[ {\left( {\pi {{\text{m}}^{ - 1}}} \right)x + \left( {5\pi {{\text{s}}^{ - 1}}} \right)t} \right]{\text{m}}$

Answer 1

To find the wave function, first recall the general equation for a wave. The wave is said to be moving in a positive direction, so apply the general equation for a wave moving in a positive direction. Using the given values find the value of wavenumber and angular frequency and put these values in the general equation. Apply the conditions given in the question to get the required wave function.

Complete step by step answer:
Given, amplitude of the wave, $A = 20\,{\text{cm}} = 0.2\,{\text{m}}$
Wavelength of the wave, $\lambda = 1\,{\text{m}}$
Velocity of the wave, $v = {\text{5 m}}{{\text{s}}^{ - 1}}$
And at $x = 0$ and $t = 0$, it has $y = 0$ and $\dfrac{{dy}}{{dt}} < 0$.

The general equation for a wave moving in positive x-direction is given by,
$y(x,t) = A\sin \left( {kx - \omega t + \phi } \right)$ (i)
where $A$ is the amplitude, $k$ is the wavenumber, $\omega$ is the angular frequency and $\phi$ is the phase of the wave.
The formula for wavenumber of a wave is,
$k = \dfrac{{2\pi }}{\lambda }$ (ii)
where $\lambda$ is the wavelength of the wave.
Here, $\lambda = 1\,{\text{m}}$ so, wavenumber of the wave is,
$k = \dfrac{{2\pi }}{1}\,{{\text{m}}^{ - 1}}$
$\Rightarrow k = 2\pi \,{{\text{m}}^{ - 1}}$
The formula for angular frequency of a wave is,
$\omega = vk$ (iii)
where $v$ is the velocity and $k$ is the wavenumber of the wave.
Here, $v = {\text{5 m}}{{\text{s}}^{ - 1}}$ and $k = 2\pi \,{{\text{m}}^{ - 1}}$so, the angular frequency of the wave is,
$\omega = 5 \times 2\pi$
$\Rightarrow \omega = 10\pi \,{{\text{s}}^{{\text{ - 1}}}}$ (iv)

Now, putting the values of $A$, $k$ and $\omega$ in equation (i), we get
$y(x,t) = 0.2\sin \left( {2\pi x - 10\pi t + \phi } \right)$ (v)
Now putting the condition $x = 0$, $t = 0$ and $y = 0$, we get
$0 = 0.2\sin \phi$
$\Rightarrow \sin \phi = 0$
$\Rightarrow \phi = 2\pi n,\,\,\,n = 0,1,2...$
Now, we differentiate equation (v) with respect to $t$ to get the value of $\dfrac{{dy}}{{dt}}$,
$\dfrac{{dy}}{{dt}} = 0.2\cos \left( {2\pi x - 10\pi t + \phi } \right) \times \left( { - 10\pi } \right)$
$\Rightarrow \dfrac{{dy}}{{dt}} = - 2\pi \cos \left( {2\pi x - 10\pi t + \phi } \right)$
At $x = 0$, $t = 0$, we have,
$\dfrac{{dy}}{{dt}} = - 2\pi \cos \left( \phi \right)$
Therefore, it satisfies the condition $\dfrac{{dy}}{{dt}} < 0$.
Putting the value $\phi = 0$ in equation (v) we get,
$y(x,t) = 0.2\sin \left( {2\pi x - 10\pi t} \right)$
$\therefore y(x,t) = \left( {0.2{\text{m}}} \right)\sin \left[ {\left( {2\pi {{\text{m}}^{ - 1}}} \right)x - \left( {10\pi {{\text{s}}^{ - 1}}} \right)t} \right]{\text{m}}$
The equation matches with option (C).

Hence the correct answer is option C.

Note: Here we have applied the general equation for a wave moving in positive direction but for a wave moving in negative direction the general equation is, $y(x,t) = A\sin \left( {kx + \omega t + \phi } \right)$. Also, while solving problems always check that the units are the same, that is all quantities are in SI units or CGS units.