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A 0.25kg is suspended in a spring by 15cm.we want to find the oscillation frequency if the mass is pulled down and then released.

1 Answer

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Answer:

8.083 rad/s ≈ 1.286 Hz

Step-by-step explanation:

The spring constant k is ...

k = Mg/l = (0.25 kg)(9.8 m/s^2)/(0.15 m) = 16 1/3 N/m

The oscillation frequency is ...

ω = √(k/M)

ω = √((49/3 N/m)/(0.25 kg)) ≈ 8.083 rad/s ≈ 1.286 Hz

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Additional comment

Since the mass is involved in both the numerator and denominator of ω, it cancels, and we have ...

ω = √(g/l) . . . . . where l is the length the mass pulls the spring from rest.

For this, we don't even need to know the spring constant or the mass.

User Macl
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