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...

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