Question

Find the maximum transverse velocity and acceleration of particles in the wire.

Answer 1

Let the wave have the form $y=Asin\left(kx\right)cos\left(\omega t\right)$

Thus transverse velocity=$\frac{dy}{dt}=-A\omega sin\left(kx\right)sin\left(\omega t\right)$

Thus maximum transverse velocity=$A\omega =0.3cm\times 2\pi \nu$

$=0.003\times 2\pi \times 60=1.13m/s$

Acceleration of particle is given by $\frac{d^{2}y}{d{t}^2}=-A{\omega }^{2}sin\left(kx\right)cos\left(\omega t\right)$

Thus maximum transverse acceleration =$A{\omega }^2=426.4m/s^2$