Question

the time period of a body executing shm is 0.05s and amplitude is 4 cm what is the max velocity

Answer 1

There are several ways to do this, depending on whether you know calculus and/or what formulas you’ve been taught. I’ll assume no calculus, but that you know the formula for the maximum potential energy of an object as it undergoes SHM:

PEmax = 1/2m(w^2)(A^2) = 1/2m(w*A)^2

m is the mass (we don’t know the mass, but we won’t need it),

w is angular frequency = 2pif =2pi/T, where T is the period

w = 2pi/.05 = 40*pi

and A is the amplitude

A = 4cm

We know that for any object, kinetic energy is KE = 1/2mV^2.

Now in SHM, the energy goes back and forth between KE and PE. And when KE is a maximum, PE=0 and when PE is a maximum, KE=0. That means

Etotal = KEmax = PEmax

So, 1/2mVmax^2 = 1/2/m(w*A)^2

Vmax^2 = (w*A)^2

Vmax = w*A

Vmax = 160pi (cm/sec)

Answer 2

Hey mate

Hope this helps you you

If the equation of motion is x(t)=A sin(2πτt)

where A = 4 cm and τ = 0.05 seconds, find vx

I will give you a hint: the maximum velocity occurs at time 0. All you have to do is find the time derivative of x, dxdt, and evaluate it at time t=0.