# Simple Harmonic Motion Problem Help

a 2.00 kg frictionless block is attached to an ideal spring with force constant 300 N/m. at t=0 the spring is neither stretched nor compressed and the block is moving in the negative direction at 12.0 m/s. Find the amplitude, the phase angle, and write and equation for the position as a function of time.

i used w=√(k/m) to find the angular frequency. i get w=12.25 rad/s

then, i'm assuming that when x=0, the cos(θ) must equal zero, in the equation x=Acos(wt+θ)

solving for θ i get θ=∏/2.

after that i used v=-wAsin(wt+θ), to find the amplitude, which i got the amplitude of 0.98m

I believe i got the phase angle and amplitude with the right concepts of simple harmonic motion, except i can't figure out how to get the right equation of x(t). i always thought x(t) was supposed to have a cos function in it, however the answer in my book says it is a sin function. If someone could explain this to me i would greatly appreciate it.

## Answers and Replies

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I believe i got the phase angle and amplitude with the right concepts of simple harmonic motion, except i can't figure out how to get the right equation of x(t). i always thought x(t) was supposed to have a cos function in it, however the answer in my book says it is a sin function. If someone could explain this to me i would greatly appreciate it.

Your results are correct. You can write the displacement during SHM in both ways: x=Acos(ωt+θ) or Asin(ωt+ψ)-they are the same with appropriate choice of the phase constants cos(ωt+θ)=sin(ωt+θ+π/2).

When the object starts from maximum displacement x=A, it is convenient to use the cosine form, as it involves zero phase constant, but the sine form is better to use when the object starts from equilibrium x=0: x=Asin(ωt).

ehild

thanks for explaining that to me!