Question

The period of oscillation of a particle executing simple harmonic motion is 4(22/7) seconds. If the amplitude of oscillation is 2.5m. calculate the maximum speed of the particle.

Answer 1

The period (T) of simple harmonic motion is the time taken for one complete oscillation. Given that the period is 4(22/7) seconds, we can use this to find the frequency (f) of the motion.

Frequency (f) = 1 / T

f = 1 / (4(22/7))
f = 7 / (4 * 22)
f = 7 / 88
f = 1 / (88 / 7)
f = 1 / 12.57
f ≈ 0.078 Hz

The frequency (f) is the number of completed oscillations per second.

The maximum speed (v_max) of the particle in simple harmonic motion is given by the equation:

v_max = 2πfA

where A is the amplitude of oscillation (2.5 m) and π is a constant.

v_max = 2πfA
v_max = 2 * (22/7) * 0.078 * 2.5
v_max = 2 * (22/7) * 0.078 * 2.5
v_max ≈ 0.982 m/s

Therefore, the maximum speed of the particle is approximately 0.982 m/s.