# Need Help at attempted problems !

1. Nov 24, 2004

### NotaPhysicsMan

:uhh: Ok any help appreciated,

First the question:

Q)A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of 3.00 Hz. The amplitude of the motion is 5.08x10^-2 m. At the point where the block has its maxiumum speed, it suddenly splits into two identical parts, only one part remaining attached to the spring. A) what is the amplitude and the frequency of the simple harmonic motion that exists after the block splits? b) repeat part a), assuming that the block splits when it is at one of its extreme positions.

This question is more conceptual I guess.

A)

Ok the max speed is when x=0.
So, when the mass splits in half, the kinetic energy will be reduced by half.
Since that is true, the potential energy is reduced by half. Since elastic potential energy is defined as Pe=1/2kx^2 and since Pe is halved. The amplitude will be smaller by square root (2) or 2^1/2. Makes sense as the mass is less, the extension is less. Ok the frequency relationship can be found by w=square root(k/m). So if the mass is halved, then w will have to be multiplied by square root(2). Makes sense as the mass is halved, oscillates more.

In part B, I don't get what difference it makes if the block broke at the extreme ends...help here? Would it really make a difference, the only thing that I see will change is that the speed is no longer at max, and x is no longer at 0.

Thanks. :tongue2:

2. Nov 24, 2004

### Staff: Mentor

You gave an excellent analysis of part a. Now do the same for part b! How do the amplitude and frequency change?

3. Nov 24, 2004

### NotaPhysicsMan

lol, yes but the part I don't understand is: what difference does it make if it were at an extreme end?

4. Nov 24, 2004

### NotaPhysicsMan

ok well let's see for A) 5.08x10^-2 x 1/square root 2, which gets 3.59 x10^ 2.
And for freq. 3.0 Hz x square root(2)= 4.24 Hz.

5. Nov 24, 2004

### Staff: Mentor

How does dropping half the mass at the extreme end affect the energy? (Is it PE or KE at that point?) Or, another way to look at it, does dropping half the mass at the extreme end affect the amplitude?

6. Nov 24, 2004

### NotaPhysicsMan

hmm, still confused, but I'll give it a crack. I'm guessing all the energy is at the extreme end has been converted to elastic potential. Since it is already at the extreme end, it has reached it's max amplitude as it would with both masses intact. So dropping half wouldn't affect?
But isn't it asking for the after effects of dropping the masses? So, if the mass is halved, the PE would be a lot less?

7. Nov 24, 2004

### Staff: Mentor

At the extreme of its displacement from equilibrium, the system has only spring potential energy--which does not depend on mass. So the energy is not affected by losing half the mass. And, it's already at maximum amplitude, so the amplitude doesn't change.

But the frequency will change.

8. Nov 24, 2004

### NotaPhysicsMan

hmm, let me see. So w=square root(k/m) and since w=2pief, I can say that if the mass is halved, then the w is multiplied by square root(2). Since w is multiplied by square root(2), then f would have to be divided by square root(2)? Say correct if I'm correct and incorrect if I'm wrong lol.

9. Nov 24, 2004

### Staff: Mentor

The frequency change is the same for both a and b. If the original frequency is $\omega_0 = \sqrt{k/m}$, replacing m by m/2 gives $\omega = \sqrt{2k/m} = \sqrt{2} \omega_0$.

10. Nov 24, 2004

### Winner

Ah ok, I wasn't thinking that w= angular frequency and thought I had to relate to another equation. Alrighty then thanks.