Question

Frequency and amplitude of a particle in simple harmonic motion A particle moves in simple harmonic motion. Knowing that the maximum velocity is 200 mm/s and the maximum acceleration is 13 m/s , determine the amplitude and frequency of the motion. Amplitude of the motion xm 1mm. Frequency of the motion勿 Hz.

Answer 1

Answer:

amplitude, a = 3.076 mm

frequency, f = 10.35 Hz

Explanation:

Vmax = 200 mm /s = 0.2 m/s

Amax = 13 m/s

Let f be the frequency and a be the amplitude.

Use the formula of maximum velocity.

Vmax = ω a

0.2 = ω a    ..... (1)

Use the formula of maximum acceleration.

Amax = ω^2 x a

13 = ω^2 a    ..... (2)

Divide equation (2) by (1)

ω = 13 / 0.2 = 65 rad/s

Put in equation (1)

0.2 = 65 x a

a = 3.076 x 10^-3 m

a = 3.076 mm

Let f be the frequency

ω = 2 π f

f = 65 / (2 x 3.14) = 10.35 Hz