Question

A simple harmonic oscillator consists of a block of mass 1.80 kg attached to a spring of spring constant 360 N/m. When t = 0.520 s, the position and velocity of the block are x = 0.200 m and v = 4.420 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

for mass attachment, in SHM,

position x = A cos (wt +phi)

velocity v = -WA sin(wt+phi)

here W = sqrt(k/m)

W = sqrt(360/1.8)

W = 14.14 rad/s

consider v/x = -W sin(wt+phi)/cos(wt+phi)

v/x = - W tan (Wt + phi)

tan (Wt+phi) = 4.42/(0.2*14.14)

Wt+phi = -1 rad

phi = -1 -(14.14*0.52)

phi = -8.35 rad

upon subsituution in 1 and 2

x = A*cos(ωt + phi)

A =0.2/cos(14.14*0.52 - 8.35)

A = 0.368 m

-----------------

b) At t = 0, x = 0.368*cos(0 - 8.35) = -0.175 m

------------------

c)

v = -ωA*sin(ωt + φ )

v = -14.14*0.368*sin(-8.35)

= -4.57 m/s

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