Question

A simple harmonic oscillator consists of a block of mass 4.10 kg attached to a spring of spring constant 240 N/m. When t = 1.70 s, the position and velocity of the block are x = 0.165 m and v = 3.780 m/s.
(a) What is the amplitude of the oscillations?

What were the (b) position and (c) velocity of the block at t = 0 s?

Answer 1

Answer:

(a) the amplitude of the oscillations is 0.52 m

(b) position of the block at t = 0 s is - 0.06 m

(c) the velocity of the block at t = 0 s is 3.96 m/s

Explanation:

given information:

m = 4.10 kg

k = 240 N/m

t = 1.70 s

x = 0.165 m

v  = 3.780 m/s

(a) the amplitude of the oscillations

x(t) = A cos (ωt+φ)

v(t) = dx(t)/dt

     =  - ω A sin (ωt+φ)

ω = $\sqrt{\frac{k}{m} }$

   = $\sqrt{\frac{240}{4.10} }$

   = 7.65 rad/s

v(t)/x(t) = - ω A sin (ωt+φ)/A cos (ωt+φ)

v(t)/x(t) = - ω sin (ωt+φ)/cos (ωt+φ)

v(t)/x(t) = - ω tan (ωt+φ)

(ωt+φ) = $tan^{-1}$ (-v/ωx)

           = $tan^{-1}$ (-3.780/(7.65)(0.165))

           = - 1.25

(ωt+φ) = - 1.25

φ = - 1.25 - ωt

   = - 1.25 - (7.65 x 1.70)

   = - 14.26

x(t) = A cos (ωt+φ)

A = x(t) / cos (ωt+φ)

   = 0.165/cos(-1.25)

   = 0.52 m

(b) position, at t=0

x(t) = A cos (ωt+φ)

x(0) = A cos (ω(0)+φ)

x(0) = A cos (φ)

      = 0.52 cos (-14.26)

      = - 0.06 m

(c) the velocity, at t=0

v(t) = - ω A sin (ωt+φ)

v(0) = - ω A sin (ω(0)+φ)

      = - ω A sin (φ)

      = - (7.65)(0.52) sin (-14.26)

      = 3.96 m/s