Question

A standing wave exist in a string of length $30\text{cm}$ which is fixed at both ends with rigid supports. The displacement amplitude of a point at a distance $10\text{cm}$ from one end is $5\sqrt{3}\text{mm}$. The distance between the two nearest point within the same loop and having displacement amplitude $5\sqrt{3}\text{mm}$ is $10\text{cm}$. Find maximum displacement amplitude of the particles in the string (in mm) upto 2 decimal places

Answer 1

Wavelength of the wave $\left(\lambda \right)=2\times 30\Rightarrow 60\text{cm}$
Let the amplitude of the wave be $=A$
Let the equation of the wave be : $y=A\sin \left(kx\right)$
Given: $x=10\text{cm}$
$y=5\sqrt{3}\text{cm}$
$\therefore 5\sqrt{3}=A\sin \left(\frac{2\pi }{\lambda }x\right)$
$5\sqrt{3}=A\sin (\frac{2\pi }{60}\times 10)$
$5\sqrt{3}=A\times \sqrt{\frac{3}{2}}$
$A=7.07\text{mm}$