Question

A horizontal stretched string fixed at both ends, is vibrating in its fifth harmonic according to the equation, $y(x,t)=\left(0.01\text{m}\right)\sin \left[\right(62.8{\text{m}}^{-1}\left)x\right]\cos \left[\right(628{\text{s}}^{-1}\left)t\right]$. Assume $\pi =3.14$, which of the following is/are the correct statements.

• A. The number of nodes is $5$
• B. The length of the string is $0.25\text{m}$
• C. The maximum displacement of the mid-point of the string from its equilibrium position is $0.01\text{m}$
• D. The fundamental frequency is $50\text{Hz}$

Answer 1

Answer: (B), (C)

The maximum displacement of the mid-point of the string from its equilibrium position is $0.01\text{m}$

String is fixed at both ends and vibrating in its fifth harmonic,



So, number of nodes $=6$
and number of loops, $n=5$

Given equation is

$y(x,t)=\left(0.01\text{m}\right)\sin \left[\right(62.8{\text{m}}^{-1}\left)x\right]\cos \left[\right(628{\text{s}}^{-1}\left)t\right]$

on comparing with, $y=2A\sin \left(kx\right)\cos \left(\omega t\right)$

$2A=0.01\text{m};k=62.8{\text{m}}^{-1};\omega =62.8{\text{s}}^{-1}$

So we get,
$2A=0.01\text{m}$

So, wavelength $\left(\lambda \right)$ of string wave

$\lambda =\frac{2\pi }{k}\Rightarrow \lambda =\frac{2\times 3.14}{62.8}=0.1\text{m}$

wave speed $\left(v\right)$

$v=\frac{\omega }{k}=\frac{628}{62.8}=10\text{m/s}$

Now, length of the string, $L=\frac{n\lambda }{2}$

$\Rightarrow L=\frac{5\times 0.1}{2}=0.25\text{m}$

Fundamental frequency, $f=\frac{v}{2L}$

Substituting the given data we get,

$f=\frac{10}{2\times 0.25}=20\text{Hz}$

Hence, options (b) and (c) are the correct alternatives.

Why this question? Let's try:- If the given string is fixed only at one end and vibrating in its $5^{th}$ harmonic then find (a) Number of nodes formed. (b) The length of the string. (c) Fundamental frequency. and compare the results obtained with the results of this question.