Question

In a sonometer experiment, the string of length $‘L’$ under tension vibrates in a second overtone between two bridges. Where is the amplitude of vibration maximum at?

Answer 1

Step 1:Given

Given that the length of the string is $L$.

Under tension vibrates in a second overtone between two bridges.

Therefore, it has two nodes and three antinodes, so we will get three maxima.

$\therefore n=3$

Let, the wavelength of the string be $\lambda$.

Step 2:: Derive the wavelength

The length of the string in terms of wavelength can be given by

$L=\frac{n\lambda }{2}$

Upon substituting the value of $n$ we get,

$L=\frac{3\lambda }{2}$

$\Rightarrow \lambda =\frac{2L}{3}$

Step 3: Calculate the three maxima

Hence, we can say that the first maximum is appearing at wavelength,

$\frac{\lambda }{4}=\frac{1}{4}\times \frac{2}{3}L$

$=\frac{L}{6}$

Similarly, second amplitude maxima will be at

$\frac{\lambda }{4}+\frac{\lambda }{2}=\frac{3}{4}\lambda$

$=\frac{3}{4}\times \frac{2}{3}L$

$=\frac{L}{2}$

And, third amplitude maxima will be at

$\frac{3}{4}\lambda +\frac{\lambda }{2}=\frac{5}{4}\lambda$

$=\frac{5}{4}\times \frac{2}{3}L$

$=\frac{5L}{6}$

Step 4: Estimate the highest maximum

Taking the LCM we can rewrite the maxima as,

$\frac{5L}{6},\frac{3L}{6},\frac{L}{6}$

Upon comparing these three maxima we can see that

$\frac{5L}{6}>$$\frac{3L}{6}>$$\frac{L}{6}$

Hence, the amplitude of vibration maximum will be at $\frac{5L}{6}$.