Question

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

Answer 1

We are asked to find the amplitude of the string at a distance $\dfrac{l}{3}$ from its one end. First recall the formula for the equation of the standing wave. Using the given conditions find the value of wavelength of the wave in the string and use this to find the amplitude of the string at the given distance.

Complete step by step answer:
Given, the length of the string is $l$, it is vibrating in its 3rd overtone and maximum amplitude is $a$. As the string is fixed at both ends, so standing waves will be formed. Equation of a standing wave is written as,
$y(x,t) = a\sin (kx)\cos (\omega t)$, where
$k$ is the propagation constant, $x$ is the distance travelled by the wave, $a$ is the maximum amplitude, $\omega$ is the angular frequency and $t$ is the time taken.
The part of the equation $2a\sin (kx)$ is the amplitude of the wave.
$A = a\sin (kx)$ (i)
Propagation constant $k$ can be written as, $k = \dfrac{{2\pi }}{\lambda }$ where $\lambda$ is the wavelength of the wave.
The formula for wavelength is written as, $\lambda = \dfrac{{2L}}{{(n + 1)}}$
Here, $n = 3$ as it is in 3rd overtone and length of the string $L = l$. Putting these values in the question for $\lambda$, we have

\Rightarrow\lambda = \dfrac{{2l}}{4}\\\ \Rightarrow\lambda = \dfrac{l}{2}$$ Now, using this value of $$\lambda $$ we find the value of propagation constant. As, $$k = \dfrac{{2\pi }}{\lambda }$$ $$ \Rightarrow k = 2\pi \left( {\dfrac{2}{l}} \right) \Rightarrow k = \dfrac{{4\pi }}{l}$$ (ii) It is given to the amplitude at a distance $$\dfrac{l}{3}$$ , that is $$x = \dfrac{l}{3}$$ Now, putting the value of $$k$$ from equation (ii) and $$x = \dfrac{l}{3}$$ in equation (i), we have $$A = a\sin (kx) = a\sin \left( {\dfrac{{4\pi }}{l}\dfrac{l}{3}} \right)$$ $$\Rightarrow A = a\sin \left( {\dfrac{{4\pi }}{3}} \right) \\\ \therefore A = - a\dfrac{{\sqrt 3 }}{2}$$ Therefore, amplitude at a distance $$\dfrac{l}{3}$$ from one end is $$\dfrac{{\sqrt 3 a}}{2}$$. **Hence, the correct answer is option C.** **Note:** A string which is stretched at both ends can oscillate up and down with different wavelengths or frequencies of vibration. But the system can oscillate with any arbitrary value of frequency, only specific values of frequencies or wavelength are allowed and these are known as normal modes.