# Question regarding angular frequency of a SHM

1. Apr 7, 2016

### Janiceleong26

1. The problem statement, all variables and given/known data

2. Relevant equations
KE=½m(ωa)2

3. The attempt at a solution
So first I did this:
2.4x10-3= ½ mω2(1.5x10-2)2
To find mω2=21.33
And substitute that into the KE eqn to find the new amplitude, which is 1.30x10-2
But I only did that because that was the only way I could think of.
My question is, why is ω constant despite changes in the max KE?

2. Apr 7, 2016

### ehild

How does the max KE depend on ω and the amplitude?

3. Apr 7, 2016

### Physics-Tutor

Linear velocity will change during the oscillation (x changes)

so the KE

will too:

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4. Apr 7, 2016

### Janiceleong26

Max KE=½m(ωa)2

5. Apr 7, 2016

### Janiceleong26

@Physics-Tutor Thanks but I'm still confused. How do we know that the angular frequency, ω, is constant?

6. Apr 7, 2016

### ehild

Changing KE means changing the amplitude if it is the same oscillator. You know that ω=√(k/m). If the spring is the same and the mass of the load is the same, why should the frequency change?

7. Apr 7, 2016

8. Apr 7, 2016

### Physics-Tutor

ehild is correct:

k is the spring constant, m the mass. These two are constant, so the angular velocity is constant.

File size:
733 bytes
Views:
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File size:
733 bytes
Views:
49
File size:
733 bytes
Views:
46
9. Apr 7, 2016

### Janiceleong26

Oh, I got it. Thanks a lot !