Question

You find a long spring and decide to create some standing waves. You note that the spring has an unstretched length of 2.0 m and a weight of 20.0 N. You anchor one end of the spring to a pold and stretch it to a length of 10.0 m. You shake the spring up and down with a period of .250 s, creating a standing wave with 4 antinodes. What is the horizontal force with which you are pulling the spring?

Answer 1

The given situation is like both ends of the spring are fixed, So the fundamental mode that can happen is with,

where L is the length of the spring and the is the wavelength of the standing waves.

When there are four antinodes formed in the stretched spring,

So, the wavelength goes down with every extra antinode formation as 1/n and so the frequency will go up as n times.

So, with four antinodes,

frequency of oscillations,

where is given such that the period of oscillations is 0.250s.

So,

This is equal to the fundamental frequency that can form in spring when it oscillates with its own mass.

In general, any spring with uniform linear mass will give rise to its effective mass as m/3.

So, the frequency can be written as,

By equating this with fundamental frequency, 1Hz, we can find out the value of k,

The extended spring length when it got stretched out is,

So, the horizontal force that is applied to pull the spring to stretch out is,

Hence the answer is 215N.

So, when there are 4 antinodes observed, the wavelenth can be estimated as follows,