Question

When a 0.770 kg mass oscillates on an ideal spring, the frequency is 1.32 Hz .

part a)

What will the frequency be if 0.270 kg are added to the original mass? Try to solve this problem without finding the force constant of the spring.

part b)

What will the frequency be if 0.270 kg are subtracted from the original mass? Try to solve this problem without finding the force constant of the spring.

Answer 1

Angular frequency is given by

w^2 = k/m

( 2 pi f ) ^2 = k / m ... (i)

( 2*3.14* 1.32)^2 = k / 0.77

k = 52.91 N/m

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a) from (i) we can easily predict that, frequency is inversely proportional to square root of mass

So

f2 / f1 = sqrt ( m1 / m2)

f2 = f1 * sqrt ( m1 / m2)

f2 = 1.32 * sqrt ( 0.77/ (0.27+ 0.77))

f2 = 1.136 Hz

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Similarly

f2 = 1.32* sqrt ( 0.77/ ( 0.77 - 0.27))

f2 = 1.638 Hz

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