Problem: An 8kg body preforms SHM of amplitude 30 cm. The restoring force is 60N, when the displacement is 30cm. Find 1. Time period 2. The acceleration, P.E and K.E When the displacement is 12 cm ( where k=f/x )

Solution:

Given:
- Mass of the body (m) = 8 kg
- Amplitude (A) = 30 cm = 0.3 m
- Restoring force (F) = 60 N
- Displacement (x) = 12 cm = 0.12 m

1. Time period (T):
The time period (T) of SHM is given by the formula:
T = 2π√(m/k)
where m is the mass of the body and k is the spring constant.

We are not given the spring constant directly, but we can calculate it using the given data:
k = F/x = 60/0.3 = 200 N/m

Substituting the values of m and k in the above formula, we get:
T = 2π√(8/200) = 0.8 s

Therefore, the time period of the SHM is 0.8 seconds.

2. Acceleration, Potential Energy, and Kinetic Energy:
The acceleration (a) of the body at any point in SHM is given by the formula:
a = -ω^2 x
where ω is the angular frequency of the SHM.

The angular frequency (ω) is given by the formula:
ω = 2π/T

Substituting the value of T in the above formula, we get:
ω = 2π/0.8 = 7.85 rad/s

Substituting the values of ω and x in the formula for acceleration, we get:
a = -ω^2 x = -(7.85)^2 * 0.12 = -7.33 m/s^2

Therefore, the acceleration of the body at a displacement of 12 cm is -7.33 m/s^2.

The potential energy (PE) of the body at any point in SHM is given by the formula:
PE = 0.5 k x^2

Substituting the value of k and x in the above formula, we get:
PE = 0.5 * 200 * (0.12)^2 = 1.44 J

Therefore, the potential energy of the body at a displacement of 12 cm is 1.44 J.

The kinetic energy (KE) of the body at any point in SHM is given by the formula:
KE = 0.5 m v^2
where v is the velocity of the body.

At the position of maximum displacement (A), the velocity of the body is zero. At the equilibrium position (E), the velocity of the body is maximum. Therefore, at a displacement of 12 cm, the velocity of the body is maximum and the kinetic energy is also maximum.

The maximum velocity (v) of the body is given by the formula:
v = ω A

Substituting the values of ω and A in the above formula, we get:
v = 7.85 * 0.

F=kx;60=k×0.3...k=200;w(angulr frqncy)=root ovr k/m=5;1)Timeperiod =2pie/w=2pie/5...f=ma..Accelertn=7.5m/s^2..AGAIN CALCULATE "w;k"AS THE DISPLACEMENT IS CHANGD...U GET.k=500;w=(500/8)^1/2 P.E=1/2mw^2x^2=3.6 J...K.E=1/2mw^2(A^2-x^2)=18.9J...my pleasure dear...