Diving board simple harmonic motion

[itryphysics]

Homework Statement

A diving board oscillates with simple harmonic mothion of frequency 3.0 cycles per second.?
What is the maximum amplitude with which the end of the board can vibrate in order that a pebble placed there to not lose contact with the board during oscillation.

Homework Equations

The Attempt at a Solution

Since the intial velocity at the bottom of a cycle is zero,
Distance = 1/2 a * t^2 where i got t by determining that quarter cycle takes
1/(4*3.0) sec. Since the accelleration exceeds that of gravity, the pebble will jump off the board. So a = 9.8 m/sec^2
Distance = 1/2 * 9.8 * (1/(4*3.0))^2
But this is the motion from the bottom to the center so the Amplitude will be twice this.
I get Amplitude = 9.8 * (1/(4*3.0))^2 = .068 meters but this answer is not correct

 

[Dick]

Distance IS NOT (1/2)*a*t^2. That's for constant acceleration problems. Oscillatory motion is NOT constant acceleration. You are using completely wrong formulas. On the good side, your idea is correct. The acceleration of the board should not exceed 9.8m/sec^2. Write the displacement of the board as x(t)=A*sin(w*t) where A is the amplitude. Find w so you get a frequency of 3 cycles per second. Now figure out x''(t), since that's acceleration, correct? Equate the maximum of that to 9.8m/sec^2.

 

[itryphysics]

Distance IS NOT (1/2)*a*t^2. That's for constant acceleration problems. Oscillatory motion is NOT constant acceleration. You are using completely wrong formulas. On the good side, your idea is correct. The acceleration of the board should not exceed 9.8m/sec^2. Write the displacement of the board as x(t)=A*sin(w*t) where A is the amplitude. Find w so you get a frequency of 3 cycles per second. Now figure out x''(t), since that's acceleration, correct? Equate the maximum of that to 9.8m/sec^2.

How do I go about finding w from 3 cycles / sec? and are you saying that x(t) = 9.8 m/s^2

 

[Dick]

How do I go about finding w from 3 cycles / sec? and are you saying that x(t) = 9.8 m/s^2

What's the period of the wave sin(wt)? And no, I said x''(t)=9.8m/s^2, x''(t) is supposed to mean the second derivative of the displacement x(t). Isn't that acceleration?

 

[itryphysics]

What's the period of the wave sin(wt)? And no, I said x''(t)=9.8m/s^2, x''(t) is supposed to mean the second derivative of the displacement x(t). Isn't that acceleration?

oh so i multiply the frequency by 2pi to get w i believe.
Im sitll confused wht do i input for "t"

 

[Dick]

Yes, I think you multiply the frequency by 2pi to get w. Your final expression for the acceleration will only have t in the form of sin(wt). You want the MAXIMUM acceleration. What's the MAXIMUM of sin(wt)?

 

[itryphysics]

Yes, I think you multiply the frequency by 2pi to get w. Your final expression for the acceleration will only have t in the form of sin(wt). You want the MAXIMUM acceleration. What's the MAXIMUM of sin(wt)?

sorry I am just confused...is it 9.8 m/s^2?

 

[Dick]

What's the maximum of sin(wt) over all t? Look at a graph of sin. Tell me what the max is. Forget this specific problem.

 

[itryphysics]

What's the maximum of sin(wt) over all t? Look at a graph of sin. Tell me what the max is. Forget this specific problem.

is it 1?

 

[Dick]

What do you mean 'is it 1?'. It IS 1. That's the correct answer. 'is it 1?' is not even an answer. It's another question. Sorry, but people have doing this to me all night. Answering an obvious question with a tentative question. Be ASSERTIVE when you know you are right.

 

[itryphysics]

What do you mean 'is it 1?'. It IS 1. That's the correct answer. 'is it 1?' is not even an answer. It's another question. Sorry, but people have doing this to me all night. Answering an obvious question with a tentative question. Be ASSERTIVE when you know you are right.

I appreciate all your help. Just take me a little further please if you can.

so i found w and set t=1

now i have 9.8= A(-sin6pi)

Is this correct?

 

[Dick]

I appreciate all your help. Just take me a little further please if you can.

so i found w and set t=1

now i have 9.8= A(-sin6pi)

Is this correct?

I'll take you as far as you want until I fall asleep, which might not be that long. The displacement is A*sin(wt) where w=6*pi/sec. I thought we had agreed to set the maximum of the second derivative of the displacement=9.8m/sec^2. What's the second derivative of sin(wt)? And whatever happened to replacing sin with it's maximum of 1? It's still there.

 

[itryphysics]

I'll take you as far as you want until I fall asleep, which might not be that long. The displacement is A*sin(wt) where w=6*pi/sec. I thought we had agreed to set the maximum of the second derivative of the displacement=9.8m/sec^2. What's the second derivative of sin(wt)? And whatever happened to replacing sin with it's maximum of 1? It's still there.

hmmm so since the max of sin is 1 , do i substitute this value right away or after i have taken the derivate of sin(wt) ?

1st derivative of sin(wt) = cos(wt)
2nd derivative = -sin(wt)

 

[Dick]

That's one problem. You aren't using the chain rule. d/dt of sin(wt) is not cos(wt). What is it? And there's a second problem. I thought we had figured out that |sin(wt)|<=1. You can set the max of the sin(wt) to 1. Don't put t=1.

 

[itryphysics]

That's one problem. You aren't using the chain rule. d/dt of sin(wt) is not cos(wt). What is it? And there's a second problem. I thought we had figured out that |sin(wt)|<=1. You can set the max of the sin(wt) to 1. Don't put t=1.

hmm i thought the w and t were numbers so i wudnt have to use the chain rule :S

and you keep saying Max of sin(wt) =1
so why are we taking teh derivative?

 

[Dick]

Ack! What is d/dt of sin(2t). If you say cos(2t), I'm going to bed right now. Look up the chain rule.

 

[itryphysics]

Ack! What is d/dt of sin(2t). If you say cos(2t), I'm going to bed right now. Look up the chain rule.

cos(2t)*(2) so cos4t

 

[Dick]

cos(2t)*2 is not equal to cos(4t). It's equal to 2*cos(2t). They aren't the same. You might be more tired than I am.

 

[itryphysics]

yes sorry i have been doing physics since the past 5-6 hrs

 

[Dick]

d/dt of sin(wt). What is it? You can nap on it, if you want to. Until we get that this will go nowhere.

 

[itryphysics]

w*cos(wt)

 

[Dick]

Yes! Second derivative of sin(wt)?

 

[itryphysics]

w^2 (-sin(wt))

 

[Dick]

Bingo. Now put it all together. The acceleration is the second derivative of A*sin(wt). You want make sure that stays less than 9.8m/s^2. What's the second derivative? What's the MAX of the second derivative? Equate that to 9.8m/s^2. Solve for A. And then we are all done.

 

[itryphysics]

MAX sin(wt)=1

A(6pi)^2 (-1) = 9.8

 

[Dick]

MAX sin(wt)=1

A(6pi)^2 (-1) = 9.8

Sure. You don't need the (-1) but I'll pass on that. The oscillation goes up and down, the sign doesn't count much for what we are doing. Just call it A(6pi)^2=9.8. I'm not going to stay up for the exciting arithmetic conclusion.

 

[itryphysics]

at last phewww . Thank you so much!