91Ó°ÊÓ

1. The movement of the bar is, d = 1.00 cm .
2. Frequency is, f = 120 Hz.
3. Linear mass density is, $\mathrm{μ}=120\mathrm{g}/\mathrm{m}$.
4. Tension is, T = 90.0 N .

We use the concept of wave motion. Using the equation of the wave, we can find its derivative to calculate speed. Using the equation of speed related to tension and linear mass density, we find the component of tension. We can find the values of displacement and energy transfer by finding the phase values. We can find energy transfer that is the power from work done and displacement. Finally, we find displacement at minimum power.

Formulae:

The general expression of the wave,

$y\left(x,t\right)=y_m\sin \left(kx-\mathrm{Ó\neg }t\right)$ (i)

The velocity of the wave,

$v=\sqrt{\frac{T}{\mathrm{μ}}}$ (ii)

The tangent angle,

$\tan \mathrm{θ}=\frac{dy}{dx}$ (iii)

The maximum rate of energy transfer in a body,

$P=\frac{W}{t}$ (iv)

Here,$y_m$ is the amplitude of the wave, k is the wave number, $\mathrm{Ó\neg }$is the angular frequency of the wave, and W is the work done.

The displacement equation is given by equation (i) is,

$y\left(x,t\right)=y_m\sin \left(kx-\mathrm{Ó\neg }t\right)$

The velocity of the wave is given by the derivative of the displacement as:

$v\left(x,t\right)=\frac{dy}{dt}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} v\left(x,t\right)=\frac{d}{dt}{y}_m\sin \left(kx-\mathrm{Ó\neg }t\right) \\ =-y_m\mathrm{Ó\neg }\cos \left(kx-\mathrm{Ó\neg }t\right) \end{gathered}$$

For the maximum value of speed, the term role="math" localid="1661166121785" $\cos \left(kx-\mathrm{Ó\neg }t\right)$should be maximum, and it is maximum at $\left(kx-\mathrm{Ó\neg }t\right)$= 0 , we get,

$v_{max}=-y_m\mathrm{Ó\neg }$ (a)

Hereis the amplitude; we can find it from the given movement of the bar as:

$$\begin{gathered} y_m=\frac{d}{2} \\ =\frac{1.00}{2} \\ =0.50\mathrm{cm} \\ =5×{10}^{-3}\mathrm{m} \end{gathered}$$

The angular frequency of the wave can be calculated as,

$\mathrm{Ó\neg }=2\mathrm{Ï€}f$

Substitute the values in the above expression, and we get,

$$\begin{gathered} \mathrm{Ó\neg }=2\left(3.14\right)\left(120\right) \\ =754\mathrm{rad}/\mathrm{s} \end{gathered}$$

The maximum speed of the wave using equation (a), we get,

$$\begin{gathered} v_{max}=\left(5×{10}^{-3}×754\right) \\ =3.77\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the maximum transverse speed is 3.77 m/s .

We can find the derivative of the displacement equation with respect to x, and we get,

$$\begin{gathered} \frac{dy}{dx}=\frac{d\left(y_m\sin \left(kx-\mathrm{Ó\neg }t\right)\right)}{dx} \\ =y_{m}k\cos \left(kx-\mathrm{Ó\neg }t\right) \end{gathered}$$

Hence, from equation (iii) and the above value of the derivative, we get

$\tan \mathrm{θ}=y_{m}k\cos \left(kx-\mathrm{Ó\neg }t\right)$

For the maximum value of $\tan \mathrm{θ}$term $\cos \left(kx-\mathrm{Ó\neg }t\right)$should be maximum, and it is maximum at$\left(kx-\mathrm{Ó\neg }t\right)=0$, we get,

$\tan \mathrm{θ}=y_{m}k$ (v)

Here we need to find the wave number k, that is,

$k=\frac{\mathrm{Ó\neg }}{v}$ (b)

Substitute the values in the equation (ii); we get the velocity of the wave as:

$$\begin{gathered} v=\sqrt{\frac{90.0}{0.120}} \\ =27.4\mathrm{m}/\mathrm{s} \end{gathered}$$

Substitute the values in equation (b); we can calculate the wave number as:

$\begin{array}{l}k=\frac{754}{27.4}\\ =27.5m^{-1}\end{array}$

Substitute the values in the equation (v); we can calculate the angle as:

So the maximum transverse component of tension can be calculated as;

$$\begin{gathered} T_{Trans}=T\sin \mathrm{θ} \\ =90.0×\sin (7.83°) \\ =12.3\mathrm{N} \end{gathered}$$

Hence, the value of transverse speed is 12.3 N .

The transverse component of tension will pull the string left; the equation for the can be written as,

$-T\left(\frac{dy}{dx}\right)=-Tk{y}_m\cos (kx-\mathrm{Ó\neg }t)$

It reaches a maximum value when$\cos (kx-\mathrm{Ó\neg }t)=-1$.

The two quantities reach their maximum values at the same value of phase. When the term$\cos (kx-\mathrm{Ó\neg }t)=-1$the value of the term$\sin (kx-\mathrm{Ó\neg }t)$will be zero, the displacement of the string y = 0 .

So, from the equation of displacement,

$y(x,t)=y_m\sin \left(kx-\mathrm{Ó\neg }t\right)$

At$\left(kx-\mathrm{Ó\neg }t\right)$= 0 the value of $\sin \left(kx-\mathrm{Ó\neg }t\right)$is zero; hence, at this value, we get transverse displacement as 0.

The tension does work when a point moves by small displacement$∆y$in time$∆t$, then the work done is,

$dW=T_{trans}×∆y$

From equation (iv), we can write the expression for energy rate transfer as,

$P=\frac{T_{trans}∆y}{∆t}$

We know that the velocity of the wave can be written as,

$\frac{∆y}{∆t}=v$

Hence, the energy transfer is given as:

$P=T_{trans}v$

For maximum power, the tension component and the speed should be maximum; we can calculate the maximum value of energy rate as:

$$\begin{gathered} P=12.3×3.77 \\ =46.4W \end{gathered}$$

Hence, the value of the rate of energy is 46.4 W .

From the above calculations, we know that maximum transfer occurs at maximum values of transverse displacement and speed, so at that value y(x,t) = 0 .

Thus, the transverse displacement y when this maximum transfer occurs is 0.

When the transverse component of tension and speed are zero, we get the minimum rate of energy transfer along the string.

Thus, the minimum rate of energy transfer along the string, when the transverse component of tension and speed are zero, is zero.

When power transfer is minimum, that is zero, the values of$\cos \left(kx-\mathrm{Ó\neg }t\right)=0$and $\sin \left(kx-\mathrm{Ó\neg }t\right)=Â\pm 1$ .

In this case, the string displacement can be calculated as,

$$\begin{gathered} y\left(x,t\right)=Â\pm y_m \\ =Â\pm 0.50\mathrm{cm} \end{gathered}$$

Hence, the value of transverse displacement is$Â\pm 0.50\mathrm{cm}$