Question

A sinusoidal transverse wave travels on a string of length $4\text{m}$ and mass $4\text{g}$. The wave speed is $20\text{m/s}$ and the wavelength is $0.2\text{m}$. If the wave is to have an average power of $50\text{W}$, then

• A. tension induced in the string is $0.4\text{N}$.
• B. frequency of wave will be $100\text{Hz}$.
• C. linear mass density of the given string must be ${10}^{-3}\text{kg/m}$.
• D. amplitude of the wave will be $11.25\text{cm}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A tension induced in the string is $0.4\text{N}$.
B frequency of wave will be $100\text{Hz}$.
C linear mass density of the given string must be ${10}^{-3}\text{kg/m}$.
D amplitude of the wave will be $11.25\text{cm}$

Given data,
Length $\left(l\right)=4\text{m}$ and wave speed $\left(v\right)=20\text{m/s}$
Mass of string $\left(m\right)=4\times {10}^{-3}\text{kg}$
Wavelength $\left(\lambda \right)=0.2\text{m}$
Linear mass density $\mu =\frac{m}{l}=\frac{4\times {10}^{-3}}{4}={10}^{-3}\text{kg/m}$
Let $T$ be the tension in the string.
From formula, $v=\sqrt{\frac{T}{\mu }}\dots \left(1\right)$
$\Rightarrow T=\mu v^2={10}^{-3}\times 20\times 20=0.4\text{N}$
Frequency $\left(f\right)=\frac{v}{\lambda }=\frac{20}{0.2}=100\text{Hz}$
Let $A$ be the amplitude of the wave.
So, average power
$P_{av}=\frac{1}{2}{\omega }^2{A}^2\mu v=\frac{1}{2}{\omega }^2{A}^2\times \frac{T}{v^2}\times v$
$\Rightarrow 50=\frac{1}{2}\times \frac{(2\pi f{)}^2{A}^2\times 0.4}{20}$
$\Rightarrow A^2=\frac{50\times 2\times 20}{0.4\times (2\pi \times 100{)}^2}=0.01266$
$\Rightarrow A\approx 0.1125\text{m}=11.25\text{cm}$
Hence, all options are correct.