Question

An object of mass $\mathbf{m}$ is attached by two stretched springs (stiffness${\mathbf{K}}_{\mathbf{s}}$ and relaxed length${\mathbf{L}}_{\mathbf{0}}$ ) to rigid walls, as shown in figure 4.60. The springs are initially stretched by an amount $\mathbf{(}\mathbf{L}\mathbf{-}{\mathbf{L}}_{\mathbf{0}}\mathbf{)}$. When the object is displaced to the right and released, it oscillates horizontally. Starting from the momentum principle, find a function of the displacement $\mathbf{x}$of the object and the time$\mathbf{t}$ describes the oscillatory motion.

(a) What is the period of the motion?

(b) If ${\mathbf{L}}_{\mathbf{0}}$ were shorter (so the springs are initially stretched more), would the period be larger, smaller, or the same?

Short answer

1. The time period is$\sqrt{2\mathrm{\pi }\left(\frac{\mathrm{m}}{2{\mathrm{k}}_{\mathrm{s}}}\right)}$ .
2. As the expression of time period is independent of length, so on increasing or decreasing the length ${\mathrm{L}}_0$, time period will not change.

Step-by-step solution

Understanding the concept

The expression for the momentum is given by,

$\mathrm{p}=\mathrm{mv}$ ……. (1)

Here $p$is the momentum, $\mathrm{m}$is the mass, $\mathrm{v}$is the velocity.

The rate of change of momentum is known as force and it is expressed as,

$\mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}}$…… (2)

Here$\mathrm{F}$ is the force.

Calculate the period of the motion.

(a)

The net force is equal to the $2{\mathrm{k}}_{\mathrm{s}}\mathrm{\omega }$,

Substitute $\mathrm{mv}$ for $\mathrm{p}$and $2{\mathrm{k}}_{\mathrm{s}}\mathrm{\omega }$for $\mathrm{F}$into the equation (2),

$\begin{array}{c}2{\mathrm{k}}_{\mathrm{s}}\mathrm{\omega }=\frac{\mathrm{d}\left(\mathrm{mv}\right)}{\mathrm{dt}}\\ =\mathrm{m}\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}\end{array}$

From the above expression,

$\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}=\frac{2{\mathrm{k}}_{\mathrm{s}}}{\mathrm{m}}\mathrm{x}$ …… (3)

Here $\mathrm{x}$ is the displacement.

Equation (3) is in the form $\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}={\mathrm{\omega }}^2\mathrm{x}$

Here$\mathrm{\omega }$ is the angular velocity,

$\begin{array}{c}{\mathrm{\omega }}^2=\frac{2{\mathrm{k}}_{\mathrm{s}}}{\mathrm{m}}\\ \mathrm{\omega }=\sqrt{\frac{2{\mathrm{k}}_{\mathrm{s}}}{\mathrm{m}}}\end{array}$

The expression for the time period is given by,

$\mathrm{T}=\frac{2\mathrm{\pi }}{\mathrm{\omega }}$

Substitute $\sqrt{\frac{2{\mathrm{k}}_{\mathrm{s}}}{\mathrm{m}}}$ for $\mathrm{\omega }$in the above equation,

$\mathrm{T}=\sqrt{2\mathrm{\pi }\left(\frac{\mathrm{m}}{2{\mathrm{k}}_{\mathrm{s}}}\right)}$

Therefore the time period is$\sqrt{2\mathrm{\pi }\left(\frac{\mathrm{m}}{2{\mathrm{k}}_{\mathrm{s}}}\right)}$ .

Dependency of time period on length.

(b)

As the expression of time period is independent of length, so on increasing or decreasing the length ${\mathrm{L}}_0$, time period will not change.