91影视

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$f_2=45\mathrm{Hz}$

Newton鈥檚 second law can be stated as, the net force ${\stackrel{\mathbf{鈫�}}{\mathbf{F}}}_{\mathbf{N}\mathbf{e}\mathbf{t}}$on a body with a mass m is related to the body鈥檚 acceleration $\stackrel{\mathbf{鈫�}}{\mathbf{a}}$by,

${\stackrel{\mathbf{鈫�}}{\mathbf{F}}}_{\mathbf{N}\mathbf{e}\mathbf{t}}\mathbf{=}\mathit{m}\stackrel{\mathbf{鈫�}}{\mathbf{a}}$

Using the basic formula for acceleration in terms of displacement and Newton鈥檚 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, $蝇=2蟿蟿f$ 鈥�(颈)

The force constant of a spring, $F=-kx$ 鈥�(颈颈)

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

$F=ma$ 鈥�(颈颈颈)

The general expression of displacement of the body,

$x-x_m\cos \left(蝇t+蠒\right)$ 鈥�(颈惫)

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:

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

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

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

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

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

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:

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

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

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

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