A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz a

Question

A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and

amplitude of 4.0 cm. If a timer is started when its displacement is maximum (hence x =

4.0 cm when t = 0), what is the speed of the mass when t = 3.0 s?

Answer: 1.2x10^-12 m/s

A mass on a spring vibrates in simple harmonic motion at a frequency of 4.0 Hz and

amplitude of 8.0 cm. If a time is started when its displacement is a maximum (hence x =

8 cm when t = 0), what is the displacement of the mass when t = 3.7 s?

Answer: 0.025 m

Please provide steps. Thank you!

Explanation / Answer

So: Æ’ = 4 Hz, A = 4 cm

and we look at the equation: x(t) = A cos (ωt + φ)
position(time) = amplitude * cos (angular frequency * time + phase constant)

φ = 0 since we're starting the motion with x (position) at it's maximum (A).

ƒ = ω/(2π) --> 4 Hz = ω / (2π) --> ω = 4 * (2π) = 8π radians/sec

Therefore we have:
x(t) = .04m * cos (8Ï€*t)

and if we plug in t=3; x(3) = .04 * cos (8Ï€ * 3)
x = .04 * cos (24Ï€) = .04 * 1 = .04 m

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