A 2.0 kg block is attached to the end of a spring with a spring constant of 350 N/m and forced to oscillate by an applied force $\mathit{F}\mathbf{\left(}\mathbf{15}\mathit{N}\mathbf{\right)}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}{\mathit{\omega }}_{\mathbf{d}}\mathit{t}\mathbf{\right)}$, where ${\mathit{\omega }}_{\mathbf{d}}\mathbf{=}\mathbf{35}\mathit{r}\mathit{a}\mathit{d}\mathbf{/}\mathit{s}$. The damping constant is $\mathit{b}\mathbf{=}\mathbf{15}\mathit{k}\mathit{g}\mathbf{/}\mathit{s}\mathbf{.}\mathit{A}\mathit{t}\mathit{t}\mathbf{=}\mathbf{0}$, the block is at rest with the spring at its rest length. (a) Use numerical integration to plot the displacement of the block for the first 1.0 s. Use the motion near the end of the 1.0 sinterval to estimate the amplitude, period, and angular frequency. Repeat the calculation for (b)${\mathit{\omega }}_{\mathbf{d}\mathbf{=}\sqrt{\raisebox{1ex}{$\mathbf{K}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{M}$}\right.}}$and (c)${\mathit{\omega }}_{\mathbf{d}\mathbf{=}}\mathbf{20}\mathit{r}\mathit{a}\mathit{d}\mathbf{/}\mathit{s}$.

• Mass,$\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{k}\mathit{g}$.
• Spring constant, $\mathit{k}\mathbf{=}\mathbf{350}\mathbf{}\mathit{N}\mathbf{/}\mathit{M}$.
• Applied force,$\mathit{F}\mathbf{=}\mathbf{\left(}\mathbf{15}\mathit{N}\mathbf{\right)}\mathbf{\left(}\mathit{s}\mathit{i}\mathit{n}{\mathit{\omega }}_{\mathbf{d}}\mathit{t}\mathbf{\right)}$.
• Angular frequency of damping,${\mathit{\omega }}_{\mathbf{d}}\mathbf{=}\mathbf{15}\mathbf{}\mathit{r}\mathit{a}\mathit{d}\mathbf{/}\mathit{s}$.
• Damping constant$\mathit{b}\mathbf{=}\mathbf{15}\mathit{k}\mathit{g}\mathbf{/}\mathit{s}\mathbf{.}$
• Time interval $\mathit{t}\mathbf{=}\mathit{O}\mathbf{}\mathit{s}\mathbf{}\mathit{t}\mathit{o}\mathbf{}\mathit{t}\mathbf{=}\mathbf{1}\mathbf{}s$.

Using the numerical integration of the damping force, we can find the amplitude of the spring. Then using the formula for damping frequency, we can find the angular frequency of the spring and using the formula of the period, we can find the period of the spring. A similar procedure can be used to find the amplitude, frequency, and period of the spring for different damping frequencies.

Formulae:

The force of damping, $f_d=-bv$where b =damping constant (i)

The angular frequency of damped oscillations,$\omega \text{'}=\sqrt{\frac{k}{m}-\frac{b^2}{4m^2}}$ (ii)

The period of damped oscillations, $T=2\pi /\omega \text{'}$(iii)

From equation (i), the displacement equation is given as follows:

Asb=(15N)(sinωdt)dt=-(15kg/s)dxdt(15N)(SINωDt)dt(15kg/s)=-dx(1m/s)(sinωdt)dt=-dx
$$\begin{gathered} Asb=\left(15N\right)(\sin {\omega }_{d}t)dt=-(15kg/s)\left(\frac{dx}{dt}\right) \\ \frac{\left(15N\right)(SIN{\omega }_{D}t)dt}{(15kg/s)}=-dx \\ (1m/s)(\sin {\omega }_{d}t)dt=-dx \end{gathered}$$

As driving frequency,${\omega }_d=15rad/s$

$$\begin{gathered} \int -dx=\int (1m/s)(\sin {\omega }_{d}t)dt \\ -\left[x\right]=(1m/s)\left[-\cos ({\omega }_{d}t\right] \end{gathered}$$

Integrating above equation between interval 0 to 1, we get the amplitude at 15 rad/s as:

$$\begin{gathered} x=(1m/ss)\left[-\cos \left(35rad\right)s+\cos \left(o\right)s\right] \\ =(1m/s)(0.903+1) \\ =(1m/s)(1.903s) \\ =1.903m \end{gathered}$$

Now using equation (ii), the angular frequency of the oscillations is given as:

$$\begin{gathered} \omega \text{'}=\sqrt{\frac{(350N/m0}{2.0kg}-\frac{{(15kg/s)}^2}{4{(2.0kg)}^2}} \\ =\sqrt{160.9375}rad/s \\ =12.686rad/s \end{gathered}$$

Using equation (iii), we can get the period of oscillation as:

$$\begin{gathered} T=\frac{2\pi }{12.686rad/s} \\ =0.495s \end{gathered}$$

A graph of x versus t is as below:

Therefore, Amplitude, period and angular frequency from interval t=0 to and ${\omega }_d=35rad/sis1.93m,0.495s,12.686rad/s$isrespectively.

We are given that the angular driving frequency is:

$$\begin{gathered} {\omega }_d=\sqrt{\frac{k}{m}} \\ =\sqrt{\frac{350N/m}{2.0kg}} \\ =13.228rad/s \end{gathered}$$

Therefore, we have from part a, we have the amplitude as:

$$\begin{gathered} x=91M/s)\left[=\cos (13.228rad)s+\left(0\right)s\right] \\ =(1m/s\left)\right(-0.788+1) \\ =0.212m \end{gathered}$$

Now, we have, angular frequency of oscillations using equation (ii) as

localid="1657282099789" $$\begin{gathered} \omega \text{'}=\sqrt{\frac{(350N/m}{2.0kg}-\frac{{(15kg/s)}^2}{4{(2.0kg)}^2}} \\ =\sqrt{160.9375rad/s} \\ =12.686rad/s \end{gathered}$$

We have, the period of oscillations using equation (iii) as:

localid="1657282109131" $$\begin{gathered} T=\frac{2\pi }{12.686rad/s} \\ =0.495s \end{gathered}$$

With the driving frequency, we have x (t) versus t as,

Therefore, Amplitude, period and angular frequency from interval t=0 to t=1 s and

${\omega }_d=\sqrt{\frac{k}{m}}=13.228rad/sis0.212m,o.495s,12.686rad/s$isrespectively.

We have been given the driving frequency as:

${\omega }_d=20rad/s$

Therefore we have from part a, the amplitude of oscillation as:

$$\begin{gathered} x=\left(1\frac{m}{s}\right)\left[-\cos \left(20rad\right)s+\cos \left(0\right)s\right] \\ =\left(1\frac{m}{s}\right)(-0.408+1) \\ 0.592m \end{gathered}$$

Now, we have, the angular frequency of oscillations using equation (ii) as:

$$\begin{gathered} \omega \text{'}=\sqrt{\frac{\left(350N/m\right)}{2.0kg}-\frac{{\left(15{\displaystyle \frac{kg}{s}}\right)}^2}{4{(2.0kg\_)}^2}} \\ =\sqrt{160.9375rad/s} \\ =12.686rad/s \end{gathered}$$

We have, the period of oscillations using equation (iii) as:

$$\begin{gathered} T=\frac{2\pi }{12.686rad/s} \\ =0.495s \end{gathered}$$

We can plot, x(t) vs t as,

Therefore,Amplitude, period and angular frequency from interval$t=0$to$t=1s$and${\omega }_d=20rad/s0.592m,0.495s,12.686rad/s$ isrespectively.