A string of length is fixed at both ends. It is vibrating in its 3^ rd overtone with maximum amplitude'a'. The amplitude at a distance / 3 from one end is

Correct answer: C. 3 a 2 For a string vibrating in its n^ th overtone (n+1)^ th harmonic) y=2 A ( (n+1) π x L ) ω t array l For .x=( )/(3)]2 A=a & n=3 y=[a ((4 π)/( ) · ( )/(3))] =a (4 π)/(3) =-a( 3 2 ) array i.e. at x=( )/(3), the amplitude is 3 a 2

BITSAT 2016

Exam: BITSAT
Year: 2016
Subject: Physics
Chapter: Waves and Sound

BITSAT 2016 · Physics · Waves and Sound

A string of length is fixed at both ends. It is vibrating in its 3^...

A string of length

\ell

is fixed at both ends. It is vibrating in its

3^{\text{rd }}

overtone with maximum amplitude'a'. The amplitude at a distance

\ell / 3

from one end is

Options

Aa
B0
C $\frac{\sqrt{3} \mathrm{a}}{2}$
D $\frac{a}{2}$

Correct answer

C. $\frac{\sqrt{3} \mathrm{a}}{2}$

Step-by-step solution

For a string vibrating in its

\mathrm{n}^{\text{th }}

overtone

(\mathrm{n}+1)^{\text{th }}

harmonic)

\mathrm{y}=2 \mathrm{ A} \sin \left(\frac{(\mathrm{n}+1) \pi \mathrm{x}}{\mathrm{L}}\right) \cos \omega \mathrm{t}

\begin{array}{l}\text{ For } \left.x=\frac{\ell}{3}\right]2 A=a & n=3 \\ y=\left[a \sin \left(\frac{4 \pi}{\ell} \cdot \frac{\ell}{3}\right)\right] \cos \omegat\\ =a \sin \frac{4 \pi}{3} \cos \oomegat=-a\left(\frac{\sqrt{3}}{2}\right) \cos \omegat \end{array}

i.e. at

x=\frac{\ell}{3},

the amplitude is

\frac{\sqrt{3} a}{2}$

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