The transverse displacement of a string (clamped at its both ends… - Vidyadip

Question

The transverse displacement of a string (clamped at its both ends) is given by
$\text{y}(\text{x, t})=0.06\sin\Big(\frac{2\pi}{3}\text{x}\Big)\cos(120\pi\text{ t})$
where x and y are in m and t in s. The length of the string is 1.5m and its mass is 3.0 × 10^–2kg.
Answer the following:
Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave?

✓

Answer

A wave travelling along the positive x-direction is given as:

$\text{y}_1=\text{a}\sin(\omega\text{t}-\text{kx})$

The wave travelling along the negative x-direction is given as:

$\text{y}_2=\text{a}\sin(\omega\text{t}-\text{kx})$

The superposition of these two waves yields:

$\text{y}=\text{y}_1+\text{y}_2=\text{a}\sin(\omega\text{t}-\text{kx})-\text{a}\sin(\omega\text{t}-\text{kx})$

$=\text{a}\sin(\omega\text{t})\cos(\text{kx})-\text{a}\sin(\text{kx})\cos(\omega\text{t})-\text{a}\sin(\omega\text{t})\cos(\text{kx})-\text{a}\sin(\text{kx})\cos(\omega\text{t})$

$=-2\text{a}\sin(\text{kx})\cos(\omega\text{t})$

$=-2\text{a}\sin\Big(\frac{2\pi}{\lambda}\text{x}\Big)\cos(2\pi\text{ vt})\ \dots(\text{i})$

The transverse displacement of the string is given as:

$\text{y}(\text{x, t})=0.06\sin\Big(\frac{2\pi}{3}\text{x}\Big)\cos(120\pi\text{ t})\ \dots(\text{ii})$

Comparing equations (i) and (ii), we have:

$\frac{2\pi}{\lambda}=\frac{2\pi}{3}$

$\therefore$ Wavelength, $\lambda=3\text{m}$

It is given that:

$120\pi=2\pi\text{v}$

Frequency, $\text{v} = 60\text{Hz}$

Wave speed, $\text{v}=\text{v}\lambda$

$=60\times3=180\text{m/s}$

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