# SHM question (Is my method correct?)

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1. Feb 13, 2017

### Faiq

1. The problem statement, all variables and given/known data
A particle P is performing simple harmonic motion with amplitude 0.25m. During each complete
oscillation, P moves with a speed that is less than or equal to half of its maximum speed for 4/3 seconds.
Find the angular frequency of P

3. The attempt at a solution
First I split 4/3 into two portions, the right and left side. Since both are symmetrical, information about left side will be same as information for right side.

Now 2/3s is the time the particle spend with velocity < 1/2 wx in one side of motion.
So
Time Spent = Time required to achieve max - time required to achieve 1/2 of max
$$\frac{2}{3} = \omega^{-1} (\sin^{-1}(\frac{Max~V}{Max~V})-\sin^{-1}(\frac{1/2~of~Max~V}{Max~V}))$$
$$\frac{2}{3} = \omega^{-1} (\sin^{-1} 1 - \sin^{-1} 0.5)$$
$$\omega = \frac{\pi}{2}$$
Note:- The answer is correct

Now the problem I realized it, even though my answer is coming correct but in my opinion, my method is not correct. The reason is I have calculated the time required by the particle to go from centre to the 1/2 speed points which makes no sense.
So can you tell me whether my method is correct and if not what is the correct method?

2. Feb 13, 2017

### BvU

Hi,

If you draw a sine and check what fraction of the period the absolute value is < 0.5 the answer is 1/3 .
So 1/3 of a period corresponds to 4/3 s. Thereby : a full period is 4 s, hence $\omega = \pi/2$ as you 'found'.
Don't agree: make a drawing.

3. Feb 13, 2017

### Faiq

How did you work out the value 1/3?

4. Feb 13, 2017

### BvU

I looked at the drawing I made

5. Feb 13, 2017

### Faiq

Î made a drawing but how did you infer the value fo 1/3?

6. Feb 13, 2017

### BvU

$[0,\pi/6] + [5\pi/6, 7\pi/6] + [11\pi/6, 2\pi]$ totals $4\pi/6$. And $4\pi/6 /(2\pi) = 1/3$

7. Feb 13, 2017

### Faiq

Oh okay thanks

8. Feb 13, 2017

### BvU

You're welcome

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