A body vibrate in simple harmonic motion with a frequency of 50 Hz and an amplitude of 4cm .find a)the period b)the acceleration at the middle and at the edge of the path of oscillation c)the veloci

period = 1/f

y = 4sin(2πf x)
v = dy/dx
find a = dv/dx at x=0 and x=pi/2
note that a(x) = -y(x)

To find the answers to your questions, we need to use the formulas for simple harmonic motion (SHM).

a) The period (T) can be found using the formula T = 1/frequency. In this case, the frequency is given as 50 Hz, so the period would be:

T = 1/50 = 0.02 seconds

b) The acceleration at the middle of the path of oscillation (x=0) in SHM is always zero. This is because the net force acting on the body is always zero at the equilibrium position.

At the edge of the path of oscillation (x = amplitude), we can find the acceleration (a) using the formula a = -ω^2 * x, where:
- ω is the angular frequency in radians per second, given by ω = 2πf (where f is the frequency).
- x is the displacement from equilibrium in meters.

First, let's convert the amplitude from centimeters to meters:
Amplitude = 4 cm = 0.04 meters.

Now, we can calculate the angular frequency (ω):
ω = 2π * 50 = 100π rad/s

Plugging these values into the formula for acceleration:

a = - (100π)^2 * 0.04 = - 40000π^2 m/s^2

So, the acceleration at the edge of the path is approximately -40000π^2 m/s^2.

c) The velocity (v) of an object in SHM can be found using the formula v = ±ω * sqrt(amplitude^2 - x^2), where:
- ω is the angular frequency in radians per second.
- x is the displacement from equilibrium in meters.

At the middle of the path of oscillation (x = 0), the velocity is always zero. This is because the velocity is maximum at the extreme positions and zero at the equilibrium position.

At the edge of the path of oscillation (x = amplitude), we can find the velocity:

v = ±ω * sqrt(amplitude^2 - x^2)
= ±100π * sqrt(0.04^2 - 0^2)
= ±100π * 0.04

So, the velocity at the edge of the path is approximately ±4π m/s.