Question

A string of length $1.5m$ with its two ends clamped is vibrating in fundamental mode. Amplitude at the centre of the string is $4mm$. Distance beween the two points having amplitude $2mm$ is

• A. $1m$
• B. $75cm$
• C. $60cm$
• D. $50cm$

Answer 1

The correct option is A $1m$

Given:

$L=1.5mA=4mm$

Equation of the wave (when time is fixed), when displacement of all parts is maximum is;
$y=4mmsin\left(\frac{\pi x}{L}\right)$
For amplitude = 2mm
$sin\left(\frac{\pi x}{L}\right)=\frac{1}{2}$
This gives us:
$\frac{x}{L}=\frac{1}{6}$ and $\frac{x}{L}=\frac{5}{6}$
The difference between these two lengths is
$\delta l=\frac{2}{3}L=1m$
Hence option A is correct.