Question

In Fig. $\mathbf{15}\mathbf{-}\mathbf{35}$, two springs are joined and connected to a block of mass 0.245 kgthat is set oscillating over a frictionless floor. The springs each have spring constant $\mathit{k}\mathbf{=}\mathbf{6430}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{m}$. What is the frequency of the oscillations?

Short answer

The frequency for the oscillations is 18.2 Hz

Step-by-step solution

The given data

1. Block of mass,$m=0.245\mathrm{kg}$
2. Spring constant,$k=6430\mathrm{N}/\mathrm{m}$

Understanding the concept of oscillatory motion

Using the combination of springs connected in a series, we can find the equivalent spring constant for the system. Next, using the formula we can find the frequency of the oscillations.

Formulae:

Angular frequency of a body under oscillation, $\omega =\sqrt{\frac{k}{m}}$(i)

The force due to spring constant,$\mathrm{F}=\mathrm{k}∆\mathrm{x}$ ….(ii)

The angular frequency of a body, $\omega =2\mathrm{\pi f}$ …….(iii)

Calculation of frequency of oscillation

As both springs are connected in a series, we can find the equivalent spring constant as given:

$k_{equivalent}=\frac{F}{Total∆x}$………..(iv)

Assuming that,

Elongation in the left spring to be$∆x_L$and Elongation in the right spring to be $∆x_R$.

As both are in a series,

Total$∆x=∆X_L+∆X_R$

Now, F is the force acting on the left force.

According to Newton’s 3^rd law, the same force will be exerted on the right spring by the left spring.

As both springs have the same spring constant and are under the same force magnitude, their elongation will also be the same i.e.$∆x_L=∆x_R$

We get

Total$∆x=2∆x$

But, from both equations (iv) & (ii), we get the value of spring constant value as:

$$\begin{gathered} k_{equivalent}=\frac{k∆x}{2∆x} \\ =\frac{k}{2} \end{gathered}$$

For the system, the angular frequency from equation (i) can be given as:

role="math" localid="1655110207339" $\omega =\sqrt{\frac{k_{equivalent}}{m}}$

Again, from equation (ii), we get the following equation as:

role="math" localid="1655110351025" $$\begin{gathered} 2\mathrm{\pi }f=\sqrt{\frac{k_{equivalent}}{m}} \\ f=\frac{1}{2\mathrm{\pi }}\times \sqrt{\frac{k}{2m}} \\ =\frac{1}{2\mathrm{\pi }}\times \sqrt{\frac{6430}{2\times 0.245}} \\ =18.2\mathrm{Hz} \end{gathered}$$

Hence, the value of frequency is 18.2 Hz.