Question

A simple harmonic oscillator consists of a block of mass 3.30 kg attached to a spring of spring constant 440 N/m. When t = 1.30 s, the position and velocity of the block are x = 0.154 m and v = 3.540 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

m = mass of the block = 3.30 kg

k = spring constant = 440 N/m

angular frequency is given as

w = sqrt(k/m) = sqrt(440/3.30) = 11.5 rad/s

x = 0.154 m

A = amplitude

velocity at any position is given as

v = w sqrt(A² - x²)

3.54 = (11.5) sqrt(A² - (0.154)²)

A = 0.344 m

b)

general equation is given as

x = A Sin(wt + )

at t = 1.30 sec

0.154 = 0.344 Sin((11.5 x 1.30) + )

= - 14.5 rad

x = A Sin(wt + )

at t = 0

x = 0.344 Sin((11.5 x 0) + (-14.5))

x = - 0.32 m

velocity is given as

v = Aw Cos(wt + )

v = (0.344 x 11.5) Cos(0 - 14.5)

v = - 1.4 m/s