- #1

- 53

- 0

For 1, using the the general SHM expression, I got x=(2 m) cos (6pi(t) + pi/2)

But for (2), plugging in 1 sec., i get 2 m. After 1 sec, shouldn't the object be back at the origin?

For (3), i would really appreciate a hint.

regards

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter redshift
- Start date

- #1

- 53

- 0

For 1, using the the general SHM expression, I got x=(2 m) cos (6pi(t) + pi/2)

But for (2), plugging in 1 sec., i get 2 m. After 1 sec, shouldn't the object be back at the origin?

For (3), i would really appreciate a hint.

regards

- #2

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 966

redshift said:

For 1, using the the general SHM expression, I got x=(2 m) cos (6pi(t) + pi/2)

But for (2), plugging in 1 sec., i get 2 m. After 1 sec, shouldn't the object be back at the origin?

For (3), i would really appreciate a hint.

regards

The amplitude of motion is 2 so the 2 multiplying cosine is correct. The intial position is t=0, x= 0 so the pi/2 gives x(0)= 2*cos(pi/2)= 0 is correct (I would have been inclined to use sin() but probably the formula you were given just uses cos()). The frequency is 3 Hz= 3 cycles per second = 6pi radians per second: Okay, (6pi)t gives you that.

Now, HOW did you get 2 m. by plugging in t= 1? 6pi(1)= pi/2= 13pi/2 and

cos(13pi/2)= 0. Did you forget to add that pi/2?

Since, as you correctly have, x= 2cos(6pi t+ pi/2), v= dx/dt= -12pi sin(6pi t+ pi/2) and a= dv/dt= -72pi

- #3

- 53

- 0

Share: