Question

A standing wave exists in a string of length $150cm$, which is fixed at both ends with rigid supports. The displacement amplitude of a point at a distance of $10cm$ from one of the ends is $5\sqrt{3}mm$. The nearest distance between the two points, within the same loop and having displacement amplitude equal to $5\sqrt{3}mm$ is $10cm$. Find the maximum displacement amplitude of the particles in the string.

• A. $20mm$
• B. $15mm$
• C. $10mm$
• D. None of these

Answer 1

The correct option is C $10mm$

Wavelength of the wave $\left(\lambda \right)=2\times 30=60cm$

let the amplitude of the wave be $=A$

let the equation of the wave be : $y=Asin\left(kx\right)$

at given $x=10cm$

$y=5\sqrt{3}cm$

$\therefore 5\sqrt{3}=Asin\left(\frac{2\pi }{\lambda }x\right)$

$5\sqrt{3}=Asin(\frac{2\pi }{60}\times 10)$

$5\sqrt{3}=A\times \frac{\sqrt{3}}{2}$

$[A=10mm]$

$\therefore$ Option (C) is correct.