Question

A 0.9 kg block attached to a spring of force constant 13.1 N/m oscillates with an amplitude of 3 cm.

A) Find the maximum speed of the block. Answer in units m/s.

B) Find the speed of the block when it is 1.5 cm from the equilibrium position. Answer in units of m/s^.

C) Find its acceleration at 1.5 cm from the equilibrium position. Answer in units of m/s².

D) Find the time it takes for the block to move from x = 0 to x= 1.5cm. Answer in units of s.

Answer 1

The position of anything undergoing simple harmonic motion could be described by

x = A sin(ωt) [1]

Velocity and acceleration can be derived by taking consecutive derivatives with respect to time.

v = Aω cos(ωt) [2]

a = -Aω² sin(ωt) [3]

For a mass-spring system,

ω² = k/m [4]

i.

Since the maximum of a cosine function is 1, from [2]

v = Aω

v = A sqrt(k/m) (from [4])

v = (0.03 m) sqrt[(13.1 N/m)/(0.9 kg)]

v = 0.114 m/s

ii.

x = A sin(ωt) [1]

0.015 m = (0.03 m) sin(ωt)

ωt = π/6

v = Aω cos(ωt) [2]

v = (0.114 m/s) cos(π/6) (from i.)

v = 0.0987 m/s

iii.

a = -Aω² sin(ωt) [3]

a = -xω² (from [1])

a = -x(k/m) (from [4])

a = -(0.03 m)[(13.1 N/m)/(0.9 kg)]

a = -0.437 m/s²