A machine part is undergoing SHM with a frequency of 5.00Hz and amplitude 1.80cm . How long does it take the part to gofrom x=0 to x= -1.80 cm

well i know that x=Acos(ωt)
and

ω=2πf so we have -1.8cm= 1.8cm * cos(2πft)

we end up with -1=cos(2πft) I know that cos(π)=-1so ft =1/2 but if that is true

then I have 5s^-1*t= 1/2 gives t= .1s

not .05s like the book says can anyone help me with the problem I'mhaving.

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Vespertilio Vespertilio · Answer 1 · 2020-01-11T15:24:36+01:00

Answer:

0.05 second

Explanation:

frequency, f = 5 H

time period is defined as the reciprocal of the frequency.

T = 1/f = 1/5 = 0.2 s

Amplitude, A = 1.80 cm

Time period is also defined as the time taken to complete one oscillation.

As the particle moves from one extreme position to one extreme position is one forth the time period of the oscillation.

As the particle goes from x = 0 to x = - 1.80 cm, so the time taken by the particle is one forth the time period of the oscillation.

t = t / 4 = 0.2 / 4 = 0.05 second.