Question

A string of length ‘ $l$ ‘ is fixed at both the ends. It is vibrating in its ${3^{rd}}$ overtone with maximum amplitude ‘ $a$ ‘. The amplitude at a distance $\dfrac{l}{3}$ from one end is:
A. $a$
B. $0$
C. $\dfrac{{\sqrt 3 }}{2}a$
D. $\dfrac{a}{2}$

Answer 1

Here, a string of length ‘ $l$ ‘ is given that is fixed at both ends. . It is vibrating in its ${3^{rd}}$ overtone and it has a maximum amplitude ‘ $a$ ‘. We will use the formula of standing waves for calculating the amplitude at one end of the string.

Complete step by step answer:
It is given in the question that a string of length $'l'$ is fixed at both the ends and it is vibrating in its ${3^{rd}}$ overtone. An overtone is defined as a frequency that is greater than the fundamental frequency.
Now, the formula used for calculating the length of $nth$ overtone of a string vibrating in $\left( {n + 1} \right)th$ is given by
$\dfrac{{\left( {n + 1} \right)\lambda }}{2} = l$
$\Rightarrow \,\lambda = \dfrac{{2l}}{{n + 1}}$
Now, as we know that
$kx \geqslant \left( {\dfrac{{2\pi }}{\lambda }} \right)x$
$\Rightarrow \,kx = \dfrac{{2\pi \left( {n + 1} \right)}}{{2l}}x$
$\Rightarrow \,kx = \dfrac{{\pi \left( {n + 1} \right)}}{l}x$
Now, the equation of the standing wave of the string is given by
$y = 2A\,\sin \left( {kx} \right)\cos \omega t$
Now, putting the value of $kx$ in the above equation, we get
$y = 2A\sin \left( {\dfrac{{\pi \left( {n + 1} \right)}}{l}x} \right)\cos \omega t$
Now, it is given in the question that the maximum amplitude of the overtone is $a$ , therefore, we will put $2A = a$ .
Now, as given in the question, number of overtone is $n = 3$
Putting these value in the expression of the standing waves, we get
$y = a\sin \left( {\dfrac{{\left( {3 + 1} \right)\pi }}{l}x} \right)\cos \omega t$
$\Rightarrow \,y = a\sin \left( {\dfrac{{4\pi }}{l}x} \right)\cos \omega t$
Now, length of the string is $x = \dfrac{l}{3}$
Now, for $x = \dfrac{l}{3}$ , the equation of standing wave is given by
$y = a\sin \left( {\dfrac{{4\pi }}{l} \times \dfrac{l}{3}} \right)\cos \omega t$
$\Rightarrow \,y = a\sin \left( {\dfrac{{4\pi }}{3}} \right)\cos \omega t$
$\Rightarrow \,y = a\left( {\dfrac{{\sqrt 3 }}{2}} \right)\cos \omega t$
Therefore, we can say that for $x = \dfrac{l}{3}$ , the amplitude of the string at one end is $\dfrac{{\sqrt 3 }}{2}a$ .
Therefore, the amplitude at a distance $\dfrac{l}{3}$ from one end is $\dfrac{{\sqrt 3 }}{2}a$

So, the correct answer is “Option C”.

Note:
An alternate method to solve the above equation is given by
$y = a\sin \left( {kx} \right)$
$\Rightarrow \,y = a\sin \left( {\dfrac{{4\pi }}{l}x} \right)$
$\Rightarrow \,y = a\sin \left( {\dfrac{{4\pi }}{l}\dfrac{l}{3}} \right)$
$\Rightarrow \,y = a\sin \left( {\dfrac{{4\pi }}{3}} \right)$
$\Rightarrow \,y = \dfrac{{\sqrt 3 }}{2}a$
Which is the required value of the amplitude of the string.