Question

what will the frequency be if 0.160 kg are subtracted from the original mass? try to solve this problem without finding the force constant of the spring.

Answer 1

Frequencies in both cases are $& f_2=1.225 \mathrm{H}_z$ and $}} \\& f_2=1.522 \mathrm{H}_z$

What is frequency ?

The frequency of a repeated event is its number of instances per unit of time. It differs from angular frequency and is sometimes referred to as temporal frequency for clarification. One event occurs per second when measuring frequency in hertz.

The quantity of waves that pass a set location in a predetermined period of time is known as frequency. Therefore, if a wave passes through in half a second, the frequency is 2 per second. The frequency is 100 times per hour if it takes 1/100 of an hour.

According to the given information

We got to know

A) the frequency of two spring

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

for Same spring

$f \propto \frac{1}{\sqrt{m}}$

So

$& \frac{f_2}{f_1}=\sqrt{\frac{m_1}{m_2}} \Rightarrow f_2=f_1 \sqrt{\frac{m_1}{m_2}}$

$& f_2=1.35 \times \sqrt{\frac{0.75}{(0.16+0.75)}} 2=222=1 \frac{1}{2}$

$& f_2=1.225 \mathrm{H}_z$

B)

$& f_2=f_1 \sqrt{\frac{m_1}{m_2}} \\& f_2=1.35 \times \sqrt{\frac{0.75}{(0.75-0.16)}} \\& f_2=1.522 \mathrm{H}_z$

Frequencies in both cases are $& f_2=1.225 \mathrm{H}_z$ and $}} \\& f_2=1.522 \mathrm{H}_z$

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