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The end point of a spring vibrates with a period of 2.1 seconds when a mass m is attached to it. When this mass is increased by 6.810×101 kg, the period is found to be 3.4 seconds. Find the value of m.

Respuesta :

Answer:

Mass attached to the spring is 41.95 kg

Explanation:

We have given time period of the spring T = 2.1 sec

Let the mass attached is m

And spring constant is k

We know that time period is given by

[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]

[tex]2.1=2\pi \sqrt{\frac{m}{k}}[/tex]---------eqn 1

Now if the mass is increased by 68.10 kg then time period become 3.4 sec

So [tex]3.4=2\pi \sqrt{\frac{m+68.10}{k}}[/tex]------eqn 2

Now dividing eqn 1 by eqn 2

[tex]\frac{2.1}{3.4}=\sqrt{\frac{m}{m+68.10}}[/tex]

[tex]0.381=\frac{m}{m+68.10}[/tex]

[tex]m=41.95 kg[/tex]

So mass attached to the spring is 41.95 kg

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