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

=========

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

========

Similarly

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

f2 = 1.638 Hz

=======

Do comment in case any doubt, will reply for sure,, good luck

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