Question

A mass m0 is attached to a spring and hung vertically. The mass is raised a short distance in the vertical direction and released. The mass oscillates with a frequency f0. If the mass is replaced with a mass nine times as large, and the experiment was repeated, what would be the frequency of the oscillations in terms of f0f0 ?

Answer 1

Answer:

The new frequency = $\frac{f_0}{3}$.

Step-by-step explanation:

Consider the provided information.

The formula for the frequency is:

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

Where k is the spring constant.

The mass oscillates with a frequency $f_0$.

$f_0=\frac{1}{2\pi}\sqrt{\frac{k}{m_0}}$

It is given that the mass is replaced with a mass nine times as large.

So replace $m_0 = 9m_0$ as shown.

$f_N=\frac{1}{2\pi}\sqrt{\frac{k}{9m_0}}$

$f_N=\frac{1}{2\pi}\times\frac{1}{3}\sqrt{\frac{k}{m_0}}\\f_N=\frac{1}{3}\times\frac{1}{2\pi}\sqrt{\frac{k}{m_0}}\\f_N=\frac{1}{3}\times f_0}}$

Here, $f_N$ represents the new frequency.

Hence, the new frequency = $\frac{f_0}{3}$.

Answer 2

The mass is replaced with a mass nine times as large. Then the new frequency is $\dfrac{f_o}{3}$.

What is the frequency?

The rate at which something occurs in a particular time period.

The formula for the frequency is given as

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

Where k is the spring constant.

The mass oscillates with a frequency $f_o$ that will be

$f_o = \rm \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m_o}}$

It is given that the mass is replaced with a mass nine times as large. Then

$f_N = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{9m_o}}\\\\\\f_N = \dfrac{1}{2\pi} * \dfrac{1}{3} \sqrt{\dfrac{k}{m_o}}\\\\\\f_N = \dfrac{1}{3} f_o$

Here, $f_N$ represents the new frequency.

Thus, the new frequency is $\dfrac{f_o}{3}$.

More about the frequency link is given below.

https://brainly.com/question/14926605