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.
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.