Question

A simple harmonic oscillator consists of a block of mass 1.80 kg attached to a spring of spring constant 210 N/m. When t = 1.70 s, the position and velocity of the block are x = 0.139 m and v = 3.550 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) 0.441 m

b) 0 m

c) 0.441 m/s

Explanation:

Simple harmonic motion equations

ω = √(k/m)

• ω - -angular velocity

• k - spring constant
• m - mass

ω  = √(210/1.8)

   = 10.8 rad/s

x = A*sin(ωt)

• x - displacement
• A - amplitude
• t - time

x = A*sin(ωt)

0.139  = A * sin(10.8 * 1.7)

A = 0.441 m

when t= 0

x = 0.441*sin(0)

x0 = 0 m

velocity =Aω*cos(ωt)

             = 0.441 * cos(0)

             = 0.441 m/s