Question

A simple harmonic oscillator consists of a block of mass 3.50 kg attached to a spring of spring constant 190 N/m. When t = 1.60 s, the position and velocity of the block are x = 0.181 m and v = 3.660 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 angular velocity is:

ω = √( k / m )
ω = √( (190 N/m) / (3.50 kg) )
ω = √62 = 7.368 rad/s

The general equation for position is
x = A sin(ωt + φ)
so
(0.181 m) = A sin((7.368 rad/s)×(1.60 s) + φ)
(0.181 m) = A sin((11.789 rad) + φ) . . . . . . . . . (1)

The general equation for velocity is
v = A ω cos(ωt + φ)
so
(3.660 m/s) = A (7.368 rad/s) cos((7.368 rad/s)×(1.60 s) + φ)
(0.497 m) = A cos((11.789 rad) + φ) . . . . . (2)

Divide equation 1 by equation 2
(0.181 m) / (0.497 m) = (A sin((11.789 rad) + φ)) / (A cos((11.789 rad) + φ))
(0.364 rad) = tan((11.789 rad) + φ)
arctan(0.364 rad) = (11.789 rad) + φ
-11.44 rad = φ

Substitute that value in equation 1
(0.129 m) = A sin((11.789 rad) + (-11.44 rad))
(0.129 m) = A sin(0.349 rad)
(0.129 m) = A×0.348
A = 0.37 m

answer (b)

Enter the right values in the equation position

x = A sin(ωt + φ)
x = (0.37 m)sin((11.789rad)×(0 s) + (-11.44 rad))
x = -0.334 m

answer (c)

Enter the right values in the equation for velocity
v = A ω cos(ωt + φ)
v = (0.37 m)×(11.789 rad/s)×cos((11.789 rad/s)×(0 s) + (-11.44 rad))
v = 1.88 m/s