Simple Harmonic Motion - Acceleration and Velocity

Peter G.

Hi,

9 (a) Write down an equation for the displacement of a particle undergoing SHM with an amplitude equal to 8.0 cm and a frequency of 14 Hz, assuming that at t = 0 displacement is = to 8.0 cm and the particle is at rest.

What I got (and this agrees with the book) was: x = 8.0 cos (28pi*t)

The problem is the book then asks for the displacement, velocity and acceleration at 0.025 seconds.

I simply plugged in the numbers for the displacement and got the -4.7 cm the book gets as well. My answers for the acceleration and velocity however, do not agree with the book. Can anyone help me spot my errors?

a = -ω²x
a = -(28pi)²x(-0.047)

I get around 36 while the book gets 360.

But, the difference in our velocity results is more significant:
v= ±ω√(xo²-x²)
v =±(28pi) √(0.08² + 0.047²)

The book gets negative -5.7ms while I get positive or negative 8.16

Thanks,
Peter G.

gneill

Looks like it could be finger problems Can you break out the intermediate steps in your calculations? Start with your expression for acceleration; what value do you calculate for (28pi)²?

Peter G.

Argh... You are right! I got 360 now. This is so creepy and embarrassing. I tried several times and always got 36... sorry Do you think I did the same thing for the velocity or I made another mistake?

gneill

Same thing, I'm afraid... more finger problems!

PeterO

Check your arithmetic, when I evaluated a = -(28pi)²x(-0.047) it gave 363.7 ??

If I leave out pi i get 36.8

Peter G.

Hi,

Sorry, was a bit busy for the last couple of hours. Yes, I managed to correct the acceleration but I am having a hard time with the velocity.

gneill

28*pi = 87.965

x_o² - x² = 0.08² - (-0.04702)² = 4.189x10^-3

87.965 * (4.189x10^-3)^(1/2) = 5.693

If you're having trouble calculating an entire expression in one go, break it up into several smaller steps where you can estimate the intermediate values and check their plausibility as you proceed.