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I have the question "The time periods are measured for a simple pendulum and a mass on a spring on Earth. The measurements are then repeated on the moon. One of the time periods is now different. Which one is it ? Explain your answer."

Is the answer that the time period on the moon is different because Gravity is now different ?

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    try looking for expression that gives the [time period of simple pendulum](https://en.wikipedia.org/wiki/Pendulum#Period_of_oscillation) and see what variables determine it. – 2017-01-01
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    You can use dimensional analysis for such questions, without the need of setting up the equations of motion or even solving them. – 2017-01-01

2 Answers 2

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The period $T$ of the pendulum depends on the local gravity $g$: $$T \approx 2\pi \sqrt{\frac{L}{g}}$$

The period of the mass on a spring does not: $$T = 2\pi \sqrt{\frac{m}{k}}$$

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Yes, because the period for small oscilations of a pendulum is given by $T= 2\pi \sqrt{\frac{l}{g}}$ where $l$ is the length of the pendulum string and $g$ is the gravitational acceleration. Hence, as in the moon the gravitational accelaration is $1,62 \frac{m}{s^2}$ and in Earth is $9.81 \frac{m}{s^2}$. Substituing these values on the equation you get smaller periods for pendulums in Earth.