Question

. A simple harmonic oscillator is of mass 0.1kg. It is oscillating with a frequency of 5/ Hz. If its amplitude of vibration is 5cm find the force acting on the particle at its extreme position. 10.50 5N 2) 2.5N 3) 3.5N 4) 4.5N

Acceleration = -Aw²
A=5cm=0.05m
f=5/pi
w=2pif
w= 2pi ×5/pi = 10
a = w²A = (10)² × 0.05 = 5m/s
force =ma = o.1 × 5 = 0.5N

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Answer 1

m = 0.1 kg

ν = 5/π Hz

A = 0.05 m

$\omega = 2\pi\nu\\k = m\omega^{2} \\F = kA\\F = 4\pi^{2} \nu^{2} mA\\F = 4*5*5*0.1*0.05\\F = 0.5 N$