# Simple Harmonic Motion

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#1
A simple pendulum of time period 1.90 s is set up alongside another pendulum of timeperiod 2.00 s. The pendulums are displaced in the same direction and released at thesame time. Calculate the time interval until they next move in phase. Explain how you arrive atyour answer. (3 marks)

I don't understand the mark-scheme of the question. Can somebody explain how to answer this question?
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6 years ago
#2
(Original post by rgtsjzfg)
A simple pendulum of time period 1.90 s is set up alongside another pendulum of timeperiod 2.00 s. The pendulums are displaced in the same direction and released at thesame time. Calculate the time interval until they next move in phase. Explain how you arrive atyour answer. (3 marks)

I don't understand the mark-scheme of the question. Can somebody explain how to answer this question?
Assume the pendulum with T=1.90s, does one oscillations, hence T=1.90s. Now how many oscillations will the other pendulum with T=2.00s have to do to have the same T=1.90s as the first pendulum?

So:

1(1.9)=y(2.00) (Where y is the number of oscillations)

It should be fairly obvious that y=1.90/2.00 oscillations.

So:

1(1.90)=(1.90/2.00)(2.00)

However we need when they will be in phase after a whole number of oscillations, we can do this via multiplying by 2.00

So: 2.00(1.90)=(1.90)(2.00)

We need to x10 to get a whole number of oscillations.

So they will next be in phase when the pendulum with T=1.90 does 20 oscillations and the pendulum with T=2.00s does 19 oscillations, it should be obvious how to calculate this time now. It is the LCM so the first time they become in phase.

SIDE-NOTE:

You could have used the idea that the pendulum with T=2.00s would have to do one less oscillation than the pendulum with T=1.90s
Hence you could have set up the equation:

(n)1.90=(n-1)2.00 (Where n is number of oscillations)

Solve for n and thus the time.
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