A simple harmonic oscillator consists of a block of mass 1.70 kg attached to a s

Question

A simple harmonic oscillator consists of a block of mass 1.70 kg attached to a spring of spring constant 144 N/m. When t = 2.00 s, the position and velocity of the block are x = +0.129 m and v = +3.415 m/s.
(a) What is the amplitude of the oscillations?
m

(b) What was the position of the mass at t = 0 s?
x = m

(c) What was the velocity of the mass at t = 0 s?
v = m/s

READ PLEASE!!! I was able to get the answer correct but would like to know why this equation is used X=Asin(wt + theta) vs X=Acos(wt + theta) for part B. Thanks

Explanation / Answer

at t=2s,

x = A*cos(wt+theta)
0.129 = A*cos(84.7*2 + theta)---(1)

v = -A*w*sin(wt+theta)

3.415 = A*(84.7)*sin(84.7*2 + theta)---(2)

(2)/(1)

tan(84.7*2+theta) = -(3.415/0.129)*84.7


theta = 82.6 rad

a) A = 0.129/cos(84.7*2 + 82.6) =

    A = 0.165 m

b) x = A*cos(w*0 + 82.6) = 0.1 m

c) v = A*w*sin(w*0 + 82.6) = 11.1 m/s

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