1. Nov 9, 2004

### cissablecat23

1a)A simple pendulum is 4.35 m long. What is the period of simple harmonic motion for this pendulum if it is hanging in an elevator that is accelerating upward at 5.90 m/s2?

b)What is the period of simple harmonic motion for this pendulum if it is placed in a truck that is accelerating horizontally at 5.90 m/s2?

2) A large block P executes horizontal simple harmonic motion as it slides across a frictionless surface with a frequency of f = 1.52 Hz. Block B rests on it, as shown in the figure below, and the coefficient of static friction between the two is μs = 0.630.

What maximum amplitude of oscillation can the system have if block B is not to slip?

1)l=4.35 m
a= 5.90 m/s/s

w^2=g/l
w^2=9.80/4.35
w^2= 2.252873563

then i have to find T.. but i don't know what formula to use...

2) w=2(pie)f
w-2(pie)(1.52 Hz)
w= 9.5504
and i don't know what else to do

2. Nov 9, 2004

### UrbanXrisis

1)

T= 2pi * sqrt(L/g)
T= 2pi* sqrt(4.35 m/ [9.8m/s^2-5.90m/s^2])

2)

"as shown in the figure below" <- picture?

3. Nov 9, 2004

### cissablecat23

it was a picture with a big block on the bottom.. a small block on the top.. with a spring attached to it..

i got the first part of the question.. it was T=2pi sqrt(l/g+a)

but i cannot get the second part of it... with it acclerating horizontally..

4. Nov 9, 2004

### UrbanXrisis

springs? I thought we were talking about pendulums?! I can't visualize #2

5. Nov 9, 2004

### cissablecat23

#2 is different from #1... it's.. a small block on a big block.. and the big block has a spring attached to it.. the coefficient of static friction was..0.630.. they want to know what's the maximum amplitude of the oscillation so that the little block doesn't slip..

f= 1.52 Hz

AND for #1.. if they change the a from upward.. to horizontal.. how does that change the answer...

6. Nov 9, 2004

### UrbanXrisis

spring attached to it where? On the bottom? sides? Need to be specific

I dont think the answer for #1 changes

7. Nov 9, 2004

### cissablecat23

i've got it.. thanks.. :)

8. Nov 20, 2004

### mikezietz

Are you sure it's 2 pi ( sqrt ( L / (g - a ) ) ? Or is it g + a?

9. Nov 21, 2004

### Staff: Mentor

In a non-accelerating frame, the period of a simple pendulum is:
$$T = 2\pi \sqrt{\frac{l}{g}}$$
In an accelerating frame, the effective "g" is different.

For case (a), the apparent acceleration is $-g\hat{y} -a\hat{y} = -(g+a)\hat{y}$, so $g_{eff} = g + a$.

For case (b), the apparent acceleration is $-g\hat{y} -a\hat{x}$, so $g_{eff} = \sqrt{g^2 + a^2}$.