Question

Please do parts (a) and part (b) below and include diagrams where applicable:

A mass m at the end of a spring vibrates with a frequency of 0.90 Hz. When an additional 520-g mass is added to m, the frequency is 0.50 Hz.

a) What is the value of m?

b) Calculate the total mass needed for the system to vibrate with a freq. of 0.45 Hz.

Answer 1

Answer:

Part a)

m = 232.1 gram

Part b)

M = 928.6 gram

Explanation:

Part a)

As we know that frequency of vibration for a given spring block system is given by formula

$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$

so if it is given as 0.90 Hz then we will have

$0.90Hz = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$

Now if additional mass is attached with it the frequency changed to 0.50 Hz

$0.50 Hz = \frac{1}{2\pi}\sqrt{\frac{k}{m + 520}}$

now divide two equations

$\frac{0.90}{0.50} = \sqrt{\frac{m + 520}{m}}$

$3.24m = m + 520$

$m = 232.1 g$

Part b)

Now the frequency is changed to 0.45 Hz

so again we will have

$0.45 Hz = \frac{1}{2\pi}\sqrt{\frac{k}{M}}$

again divide it with first equation above

$\frac{0.90}{0.45} = \sqrt{\frac{M}{m}}$

as we know that m = 232.1 g

so total mass needed for 0.45 Hz will be

$M = 928.6 gram$