Question

A simple harmonic oscillator consists of a block of mass 3.4 kg attached to a spring of spring constant 120 N/m. When t = 0.84 s, the position and velocity of the block are x = 0.127 m and v = 3.23 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

a)

m = mass of the block = 3.4 kg

k = spring constant = 120 N/m

angular frequency of the spring block system is given as

w = sqrt(k/m)

w = sqrt(120/3.4)

w = 5.94 rad/s

let the position at any time is given as

x(t) = A Cos(wt + )

where

A = amplitude

w = angular frequency

= time constant

t = time

At t = 0.84 ,

x(0.84) = A Cos((5.94)(0.84) + )

0.127 = A Cos((5.94)(0.84) + ) Eq-1

Equation for velocity is given as

v(t) = - A w Sin(wt + )

at t = 0.84

3.23 = - A (5.94) Sin((5.94)(0.84) + ) Eq-2

Dividing eq-2 by eq-1

3.23/0.127 = - A (5.94) Sin((5.94)(0.84) + ) /(A Cos((5.94)(0.84) + ))

3.23/0.127 = - (5.94) tan((5.94)(0.84) + )

- 4.3 = tan((5.94)(0.84) + )

1.34 = (5.94)(0.84) +

= - 3.65 rad

Using eq-1

0.127 = A Cos((5.94)(0.84) + )

0.127 = A Cos((5.94)(0.84) + (- 3.65))

A = 0.554 m

b)

position at any time is given as

x(t) = A Cos(wt + )

at t = 0

x(0) = A Cos(w(0) + )

x(0) = (0.554) Cos(w(0) + (- 3.65))

x(0) = - 0.484 m

velocity at any time is given as

v(t) = - Aw Sin(wt + )

at t = 0

v(0) = - Aw Sin(w(0) + )

v(0) = - (0.554)(5.94) Sin(w(0) + (- 3.65))

v(0) = - 1.6 m/s