Question

Consider a spring that exerts the following restoring force:

\- kx{\text{ for }}x > 0{\text{ }} \\\ \- 2kx{\text{ for }}x < 0 \\\ \end{aligned} \right\\}$$ A mass on a frictionless surface is attached to this spring, displaced to $$x = A$$ by stretching the spring and released. (a) Find the period of motion. (b) What is the most negative value of x that the mass m reaches?

Answer 1

The time period of motion of a spring is the time taken by the system to complete one oscillation. It depends both on the mass of the system and the spring constant.
Formula used:$T = 2\pi \sqrt {\dfrac{m}{k}}$, where T is the time period of oscillation, m is the mass attached on the spring and k is the spring constant.

Complete step by step answer
We are provided with the restoring force value when the spring moves about the mean position. We know that time period is defined as the time taken to one complete cycle and hence, in this case the time period will be the sum of the time due to the two restoring forces. We know that:
$T = 2\pi \sqrt {\dfrac{m}{k}}$
On comparing with the general equation of $F = -kx$, we can see that the two spring constants are k and 2k. Hence, the time period will be:
$T = \dfrac{1}{2}\left( {2\pi \sqrt {\dfrac{m}{k}} + 2\pi \sqrt {\dfrac{m}{{2k}}} } \right)$
We multiplied them by a half, because these give the time period for half of the oscillation. Solving it further:
$T = \dfrac{1}{2} \times 2\pi \sqrt {\dfrac{m}{k}} \left( {1 + \dfrac{1}{{\sqrt 2 }}} \right) = \pi \sqrt {\dfrac{m}{k}} \times \dfrac{{\sqrt 2 + 1}}{{\sqrt 2 }}$
Hence, the time period of motion will be:
$T = \pi \sqrt {\dfrac{m}{{2k}}} (\sqrt2 + 1)$
To find the displacement for which the mass on the spring reaches a certain value, we use the energy conservation equation. We know that the energy of a spring is given as:
$E = \dfrac{1}{2}k{x^2}$
This energy will be equal to the energy possessed by the spring when the spring is displaced to A:
$\dfrac{1}{2}k{A^2} = \dfrac{1}{2}(2k){x^2}$
Solving for x, we get:
${A^2} = 2{x^2}$
$x = \pm \dfrac{A}{{\sqrt 2 }}$

Hence, the most negative value will be $x = - \dfrac{A}{{\sqrt 2 }}$

Note: The restoring force in the spring is responsible for making it go back to its original size. As we saw, it is only a function of the displacement of the spring from its equilibrium position, and has a negative sign due it being responsible for the spring going back to its position.