Question

In Figure 15-31, two springs are attached to a block that can oscillate over a frictionless floor. If the left spring is removed, the block oscillates at a frequency of 30 Hz. If, instead, the spring on the right is removed, the block oscillates at a frequency of 45 Hz. At what frequency does the block oscillate with both springs attached?

Short answer

The block will oscillate at the frequency54 Hz with both springs attached.

Step-by-step solution

The given data

1) When the left spring is removed block oscillates at $f_1=30\mathrm{Hz}$

2) When the right spring is removed block oscillates at$f_2=45\mathrm{Hz}$

Understanding the concept of Newton’s second law of motion

Newton’s second law can be stated as, the net force ${\overrightarrow{\mathbf{F}}}_{\mathbf{N}\mathbf{e}\mathbf{t}}$on a body with a mass m is related to the body’s acceleration $\overrightarrow{\mathbf{a}}$by,

${\overrightarrow{\mathbf{F}}}_{\mathbf{N}\mathbf{e}\mathbf{t}}\mathbf{=}\mathit{m}\overrightarrow{\mathbf{a}}$

Using the basic formula for acceleration in terms of displacement and Newton’s 2nd law, we can findin terms the spring constant of the given two springs. From this, we can find the frequency for oscillations created by each spring, and after adding them, we can get the total frequency of the system.

Formula:

The angular frequency of an oscillation, $\omega =2\tau \tau f$ …(i)

The force constant of a spring, $F=-kx$ …(ii)

The force of a body in motion due to Newton’s law of motion,

$F=ma$ …(iii)

The general expression of displacement of the body,

$x-x_m\cos \left(\omega t+\varphi \right)$ …(iv)

Calculation of oscillation frequency

When the block is displaced fromthe equilibriumposition, the net force exerted on the left and right springs attached to the block using equation (ii) is given by:

$$\begin{gathered} F_1=-k_{1}x \\ {F}_2=-k_{2}x \\ {F}_{Net}=F_1+F_2 \\ =-k_{1}x-k_{2}x.......................\left(a\right) \end{gathered}$$

According to equation (iii), we get the equation (a) as:

$ma=-k_{1}x-k_{2}x$

$m\times \left(\frac{d^{2}x}{d{t}^2}\right)=-k_{1}x-k_{2}x\left(\because a=\frac{d^{2}x}{d{t}^2}\right)$

role="math" localid="1657270444191" $$\begin{gathered} m\times \left(\frac{d^2\left(x_m\cos \left(\omega t+\varphi \right)\right)}{d{t}^2}\right)=-k_1\left(x_m\cos \left(wt+\varphi \right)\right)-k_2\left(x_m\cos \left(\omega t+\varphi \right)\right) \\ m\times \frac{d\left(\omega x_m\times -\sin \left(\omega t+\varphi \right)\right)}{dt}=-x_m\cos \left(\omega t+\varphi \right)\times \left(k_1+k_2\right) \\ m\times {\omega }^2{x}_m\cos \left(\omega t+\varphi \right)=-x_m\cos \left(\omega \varphi \right)\times \left(k_1+k_2\right) \\ m{\omega }^2=k_1+k_2 \\ {\omega }^2=\frac{k_1+k_2}{m} \\ \omega =\sqrt{\left(\frac{k_1+k_2}{m}\right)} \end{gathered}$$

Using equation (i) and the given values, we get the frequency of oscillation as:

$f=\frac{1}{2\mathrm{\pi }}\times \sqrt{\left(\frac{k_1+k_2}{m}\right)}$

The single spring acting along will perform a simple harmonic motion and frequency is given to us for both sides of springs can be written as:

$$\begin{gathered} f_1=\frac{1}{2\mathrm{\pi }}\times \sqrt{\left(\frac{k_1}{m}\right)} \\ =30\mathrm{Hz} \\ {\mathrm{f}}_2=\frac{1}{2\mathrm{\pi }}\times \sqrt{\left(\frac{{\mathrm{k}}_2}{\mathrm{m}}\right)} \\ =45\mathrm{Hz} \end{gathered}$$

The frequency at which the block will oscillate with both springs attached will be

$$\begin{gathered} f=\sqrt{f_1^2+f_2^2} \\ =\sqrt{{\left(30Hz\right)}^2+{\left(45Hz\right)}^2} \\ =\sqrt{900H{z}^2+2025H{z}^2} \\ =\sqrt{2925H{z}^2} \\ =54.08Hz \\ \approx 54Hz \end{gathered}$$

Hence, the value of frequency is found to be$54\mathrm{Hz}$ .