A500g particle oscillating in shm travels 30cm between the two extreme points in its motion with an average speed of 100cm/s.

Answers:
a. angular frequency = 5/3 Hz
b. maximum force = -8.216 * 10^3 newton
c. maximum speed = 1.56999 m/sec

Explanation:
Part a:
We are given that the particle oscillates 30 cm with velocity of 100 cm/sec
The rule that relates velocity, distance and time is as follows:
velocity = distance / Δtime
Now, substitute with the givens in the above equation to get the time as follows:
100 = 30/Δtime
Δtime = 30/100 = 0.3 sec
Now, we have the period = 2Δt = 2*0.3 = 0.6 seconds
Now, the frequency is the inverse of the period. This means that:
frequency = 1/t = 1/0.6 = 5/3 Hz

Part b:
In the simple harmonic motion, the period can be calculated as follows:
T = 2π√(m/k)
where:
T is the period = 0.6 sec
m is the mass of the particle = 500 gram
Substitute to get k as follows:
0.6 = 2(3.14)√(500/k)
15/157 = √(500/k)
225/24649 = 500/k
k = 54.7755 * 10^3
Now, the distension in the spring = 30/2 = 15 cm = 0.15 m
(This is because the two end points are separated equally)
Finally, we will use Hook's law to get the force as follows:
F = -kx
F = -54.7755 * 10^3 * 0.15 = -8.216 * 10^3 newton
This force is considered the maximum force is the distension is maximum

Part c:
Velocity is maximum when the acceleration is zero. This can happen when the distension in also zero (distension is balanced).
This means that the maximum velocity occurs when there is no restoring force exists. This means that we have only:
i. kinetic energy in the ends
ii. elastic potential energy
These two energies must be equal, therefore:
KE = PE
0.5mv^2 = 0.5kx^2
mv^2 = kx^2
500 * v^2 = (54.7755 * 10^3) * (0.15)^2
v^2 = 2.4648
v = 1.56999 m/sec

Hope this helps :)

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