Question

An object of mass 0.2 kg is undergoing SHM of frequency 6 Hz and amplitude of 0.25 m. Find its period, frequency, angular frequency, and maximum speed. a) Period = 0.167 s, Frequency = 6 Hz, Angular frequency = 37.7 rad/s, Maximum speed = 0.25 m/s b) Period = 0.167 s, Frequency = 6 Hz, Angular frequency = 113.1 rad/s, Maximum speed = 0.25 m/s c) Period = 0.333 s, Frequency = 3 Hz, Angular frequency = 18.85 rad/s, Maximum speed = 0.25 m/s d) Period = 0.333 s, Frequency = 3 Hz, Angular frequency = 37.7 rad/s, Maximum speed = 0.25 m/s

Answer 1

The period of an object undergoing simple harmonic motion can be calculated by taking the inverse of its frequency. The angular frequency can be calculated by multiplying the frequency by 2π, and the maximum speed is equal to the angular frequency multiplied by the amplitude.

Step-by-step explanation:

Given:

Mass (m) = 0.2 kg

Frequency (f) = 6 Hz

Amplitude (A) = 0.25 m

To find: Period (T), Angular frequency (ω), and Maximum speed (v)

Calculating Period (T):

T can be calculated using the formula:

T = 1/f

Substituting the given frequency into the formula:

T = 1/6 = 0.167 seconds

Calculating Angular frequency (ω):

ω can be calculated using the formula:

ω = 2πf

Substituting the given frequency into the formula:

ω = 2π(6) = 37.7 rad/s

Calculating Maximum speed (v):

Maximum speed is equal to the angular frequency multiplied by the amplitude:

v = ωA

Substituting the given values into the formula:

v = (37.7 rad/s)(0.25 m) = 9.425 m/s

Therefore, the correct answer is:

a) Period = 0.167 s, Frequency = 6 Hz, Angular frequency = 37.7 rad/s, Maximum speed = 0.25 m/s