# When a 0.850 kg mass oscillates on an ideal spring, the frequency is 1.43 Hz.What is the frequency if 0.320 kg is added to the original mass and when 0.32 kg is subtracted from the original mass?

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### 1 Answer

For a spring we have the frequency angular frequency W^2 = k/m where m is the mass on the spring and k is the spring constant.

When a mass of 0.85 kg is attached to the spring the frequency is 1.43 Hz.

=> 1.43^2 = k/(0.85)

To find the change in frequency for a change in mass, we have to use the value of k which can be derived from the equation given above.

k = 1.43^2*0.85

For an increase in mass of 0.32 kg:

W = sqrt [1.43^2*0.85/(0.85 + 0.32)]

=> 1.218 Hz

For a decrease in mass of 0.32 kg:

W = sqrt [1.43^2*0.85/(0.85 - 0.32)]

=> 1.810 Hz