Question

A spring stretches by 0.0177 m when a 2.82-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 7.42 Hz?

Answer 1

The mass attached to the spring must be 0.72 kg

Explanation:

The frequency of vibration of a spring-mass system is given by:

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

where

k is the spring constant

m is the mass attached to the spring

We can find the spring constant by using Hookes' law:

$F=kx$

where

F is the force applied on the spring

x is the stretching of the spring

When a mass of m = 2.82 kg is applied to the spring, the force applied is the weight of the mass, so we have

$mg=kx$

and using $g=9.8 m/s^2$ and $x=0.0177 m$, we find

$k=\frac{mg}{x}=\frac{(2.82)(9.8)}{0.0177}=1561.3 N/m$

Now we want the frequency of vibration to be

f = 7.42 Hz

So we can rearrange eq.(1) to find the mass m that we need to attach to the spring:

$m=\frac{k}{(2\pi f)^2}=\frac{1561.3}{(2\pi (7.42))^2}=0.72 kg$

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