Question

A standing wave is created on a string of length $120\text{m}$ and it is vibrating in $6^{th}$ harmonic. Maximum possible amplitude of any particle is $10\text{cm}$ and maximum possible velocity will be $10\text{cm/s}$. Which of the following statement(s) is/are correct?

• A. Angular wave number $\left(k\right)$ of the wave will be $\frac{\pi }{20}{\text{cm}}^{-1}$.
• B. Angular wave number $\left(k\right)$ of the wave will be $\frac{\pi }{10}{\text{cm}}^{-1}$.
• C. Time period of any particles SHM will be $4\pi \text{sec}$.
• D. Time period of any particles SHM will be $2\pi \text{sec}$.

Answer 1

Answer: (A), (D)

The correct options are
A Angular wave number $\left(k\right)$ of the wave will be $\frac{\pi }{20}{\text{cm}}^{-1}$.
D Time period of any particles SHM will be $2\pi \text{sec}$.

Given that,
Length of string $l=120\text{m}$
Maximum amplitude $A=10\text{cm}$
Maximum speed $v_{max}=10\text{cm/s}$
Now,
$l=\frac{n\lambda }{2}$
$6\left(\frac{\lambda }{2}\right)=120\text{cm}$
$\Rightarrow \lambda =40\text{cm}$
$\Rightarrow k=\frac{2\pi }{\lambda }$
$\Rightarrow k=\frac{2\pi }{40}$
$\Rightarrow k=\frac{\pi }{20}{\text{cm}}^{-1}$
Also
$A\omega =v_{max}$
$\omega =\frac{v_{max}}{A}=\frac{10\text{cm/s}}{10\text{cm}}$
$\omega =1$
Time period of oscillation is
$\Rightarrow T=\frac{2\pi }{\omega }=2\pi \text{sec}$.
Thus, options (a) and (d) are correct.