Question

An object is undergoing SHM with period 0.810sand amplitude 0.320 m. At t = 0, the object is at x= 0.320 m and is instantaneously at rest.

Part A

Calculate the time it takes the object to go from x = 0.320 m, to x = 0.160 m.

Express your answer with the appropriate units.

Part B

Calculate the time it takes the object to go from x = 0.160 m, to x = 0.

Express your answer with the appropriate units.

Answer 1

we can write the equation of motion for a harmonic oscillator as

x(t) = A cos[(2 pi/P) t] where P is the period, t is time, and A is the amplitude

for this oscillator we have

x(t) = 0.32 cos[(2 pi/0.810) t]

we know t = 0 when x=0.32, so we find the value of t when x =0.16m:

0.16 = 0.32 cos[2 pi/0.810) t]

0.5 = cos[(2pi/0.810) t]

pi/3 = (2 pi/0.810)t => t= 0.13s

it takes 0.13 s to go from 0.32 to 0.16m

the time at which x=0 is:

0 = 0.32 cos[2 pi/0.810)t]

pi/2 = (2 pi/0.810)t

t = 0.195s

so it takes 0.195s-0.130s = 0.065s to go from 0.16 to 0 m