A mass m at the end of a spring vibrates with a frequency of 0.88 Hz When an additional 680g mass is added to m the frequency is 0.60 Hz what is the value of m?

Answer 1

By Master Amir (UPSI)

1) Let k = force constant of the spring,

M = mass attached to the end of the spring,

f = frequency

f = (1/2pi)*sqrt(k/M)

When M = m, then f = 0.88 Hz

Therefore,

0.88 = (1/2pi)*sqrt(k/m)-----------------------...

When M = m+680 g = m + 0.68 kg, then f = 0.60 Hz

Therefore,

0.60 = (1/2pi)*sqrt(k/(m + 0.68))----------------(2)

Dividing (1) by (2):-

0.88/0.60 = sqrt(k/m)/sqrt(k/(m + 0.68))

Or 1.47 = sqrt((m + 0.68)/m)

Or 1.47 = sqrt(1 + 0.68/m)

Taking square on both sides: -

2.16 = 1 + 0.68/m

Or 0.68/m = 2.16 - 1

Or 0.68/m = 1.16

Or m = 0.68/1.16

Or m = 0.586 kg = 586 g

Ans: 586 g