Question

A string is in horizontal position as shown in figure, linear mass density of string is $\frac{4}{15}{\text{kg m}}^{-1}$. Find the length $\left(\text{in m}\right)$ of string if a standing wave is set on string with two loops formed. The frequency of the wave is $5\text{Hz}$. (Use $g=10{\text{m/s}}^2$)

Answer 1

For the pulley system


Let $a$ be the acceleration of $2\text{kg}$ block, then

$a=\left(\frac{m_2-m_1}{m_1+m_2}\right)g$

$\Rightarrow a=\left(\frac{2-1}{1+2}\right)g=\left(\frac{g}{3}\right)$

FBD of $1\text{kg}$ mass,


$\Rightarrow T-mg=ma$

$\Rightarrow T=m(g+a)$

$\Rightarrow T=g+\frac{g}{3}=\left(\frac{4g}{3}\right)$

Tension in the horizontal string will be $2T$

$\Rightarrow T_h=2T=\frac{8g}{3}$

Velocity of wave,

$V_w=\sqrt{\frac{T_h}{\mu }}=\sqrt{\frac{8g}{3\left(\frac{4}{15}\right)}}=\sqrt{10g}$

$\Rightarrow V_w=10{\text{ms}}^{-1}$

Wave length, $\lambda =\frac{V}{f}$

$\Rightarrow \lambda =\frac{10}{5}=2\text{m}$

Given that, there are two loops in standing wave, so the length of the string will be equal to one wavelength.

Length of string, $L=\lambda =2\text{m}$

Hence, $2$ is the correct answer.

Why this question? This gives clear understanding about waves and also help in revising newton laws of motion. It also helps in remembering the following formulae, $\lambda =V_{m}T,V_m=\sqrt{\frac{T}{\mu }}$