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An object with mass m in simple harmonic motion on a vertical spring is observed for 5 full oscillations. The time is measured to be 13 seconds.

What is the angular frequency (3dp)? A previous experiment using a test mass shows the spring constant is 24N/m. What is the mass of the object (3dp)?

This is the question I need to answer, I don't know what is going on from my notes. I don't have equations for these quantities with clearly defined terms. Could you please tell me what I need to know to figure this out?

Penny
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1 Answers1

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In such a spring-mass system, the time period $T$ of one oscilation is related to the mass $m$ of the object and the spring constant $k$ by the equation

$$T = 2\pi \sqrt{\frac{m}{k}}$$

(https://en.wikipedia.org/wiki/Simple_harmonic_motion#Mass_on_a_spring)

You know $T$ and $k$ from the text, now search for $m$.

Also, the angular frequency $\omega$ is related to the frequency $f$ by

$$\omega = 2\pi f $$

(https://en.wikipedia.org/wiki/Angular_frequency)

Whereas the period $T$ and the frequency $f$ are related by

$$f = \frac{1}{T}$$

That should be everything you need.

Maximilian Gerhardt
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  • This is the first thing about this physics work that has made sense to me. Just to verify, the angular frequency is ${{2\pi } \over {13}} \approx 0.48rad/s$? I get the formula for the mass as $m = {{k{T^2}} \over {4{\pi ^2}}}$ leading to a mass of 102.74kg which seems very high... – Penny Apr 29 '16 at 14:08
  • They say they observed 5 oscilations within 13 seconds, wouldn't that make $T = 13/5 = 2.6 \mathrm{ [sec]}$? Then the mass would be $\approx 4.11 \mathrm{ kg}$. – Maximilian Gerhardt Apr 29 '16 at 14:11
  • Oh, of course! I thought I'd made a silly mistake. – Penny Apr 29 '16 at 14:20