Question

A mass m at the end of a spring vibrates with a frequency

Answer 1

Answer:

m = 0.59 kg.

Explanation:

First, we need to find the relation between the frequency and mass on a spring.

The Hooke's law states that

$F = -kx$

And Newton's Second Law also states that

$F = ma = m\frac{d^2x}{dt^2}$

Combining two equations yields

$a = -\frac{k}{m}x$

The term that determines the proportionality between acceleration and position is defined as angular frequency, ω.

$\omega = \sqrt{\frac{k}{m}}$

And given that ω = 2πf

the relation between frequency and mass becomes

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

Let's apply this to the variables in the question.

$0.88 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\\0.60 = \frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}\\\frac{0.88}{0.60} = \frac{\frac{1}{2\pi}\sqrt{\frac{k}{m}}}{\frac{1}{2\pi}\sqrt{\frac{k}{m+0.68}}}\\1.4667 = \frac{\sqrt{m+0.68}}{\sqrt{m}}\\2.15m = m + 0.68\\1.15m = 0.68\\m = 0.59~kg$