Question

A standing wave pattern on a string is described by $\mathit{y}\mathbf{(}\mathit{x}\mathbf{,}\mathit{t}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{040}\mathbf{(}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{5}\mathbf{\pi x}\mathbf{)}\mathbf{(}\mathbf{cos}\mathbf{}\mathbf{40}\mathbf{\pi t}\mathbf{)}$, where x and y are in meters and is in seconds. For, x > 0 what is the location of the node with the (a) Smallest, (b) Second smallest, and (c) Third smallest value of x? (d) What is the period of the oscillatory motion of any (non-node) point? What are the (e)Speed and (f) Amplitude of the two traveling waves that interfere to produce this wave? For t > 0, what are the (g) First, (h) Second, and (i) Third time that all points on the string have zero transverse velocity?

Short answer

1. Location of node withthesmallest value is at 0 m
2. Location of node with the second smallest value is at 0.20 m
3. Location of node withthethird smallest value is at 0.40 m
4. Period of the oscillatory motion of any point is 0.05 s
5. Speed of the two travelling waves that interfere to produce this wave is 8.0 m/s
6. Amplitude of the two travelling waves that interfere to produce this wave is 0.020 m
7. First time that all points on the string have zero transverse velocity is at 0 s
8. Second time that all points on the string have zero transverse velocity is at 0.025 s
9. Third time that all points on the string have zero transverse velocity is 0.05 s

Step-by-step solution

Given data

$$\begin{gathered} \mathrm{Equation}\mathrm{of}\mathrm{the}\mathrm{standing}\mathrm{wave},y(x,t)=0.040\left(\sin 5\mathrm{\pi x}\right)(\cos 40\mathrm{\pi }t) \\ \mathrm{Angular}\mathrm{frequency},\omega =40\mathrm{\pi } \\ \mathrm{Wavenumber},k=5\mathrm{\pi } \end{gathered}$$

Understanding the concept of node

We can find the location of the nodes with different values of x from the condition that the amplitude of the wave must be zero at that point. We can get the angular speed and wave number and amplitude of the given wave by comparing the given equation of the wave with the general wave equation. Then, from this, we can find the period, speed of the standing wave, and amplitude of the interfered wave using the corresponding relations. From the given displacement of the wave, we can find the expression for its speed. Then, inserting the given conditions we can find the answers for parts g), h) and i).

Formula:

The angular frequency of the wave, $\omega =2\mathrm{\pi f}..............\left(1\right)$

The period of an oscillation,$T=\frac{1}{f}............\left(2\right)$

The velocity of a wave,$V=\frac{\omega }{k}............\left(3\right)$

The general expression of the wave, $y=y_m\sin (kx-\omega t).....\left(4\right)$

Step 3(a): Calculation of location of node with the smallest value

The standing wave equation is given as:

$y(x,t)=0.040\left(\sin 5\mathrm{\pi x}\right)\left(\cos 40\mathrm{\pi }t\right)$

Now the nodes will be located where

$\sin 5\mathrm{\pi x}=0$

For the value of thefunction to be zero, the value for$5\mathrm{\pi x}$must be in the range of$0,\mathrm{\pi },2\mathrm{\pi },3\mathrm{\pi }........$

So, from this we can say that,

Location of the node with the smallest value is given at x=0:

$$\begin{gathered} 5\mathrm{\pi x}=0 \\ \mathrm{x}=\frac{0}{5\mathrm{\pi }} \\ =0 \end{gathered}$$

Hence, the node is located at 0 m

Step 4(b): Calculation of location of node with second smallest value

From, the give range, the location of the node with the second smallest value for x is given as:

$$\begin{gathered} 5\mathrm{\pi x}=\mathrm{\pi } \\ 5\mathrm{x}=1 \\ \mathrm{x}=\frac{1}{5} \\ =0.20\mathrm{m} \end{gathered}$$

Hence, the node is located at 0.20 m

Step 5(c): Calculation of location of node with third smallest value

From, the give range, the location of the node with the third smallest value for x is given as:

$$\begin{gathered} 5\mathrm{\pi x}=2\mathrm{\pi } \\ 5\mathrm{x}=2 \\ \mathrm{x}=\frac{2}{5} \\ =0.40\mathrm{m} \end{gathered}$$

Hence, the location of node is 0.40 m

Step 6(d): Calculation of period of oscillation at any point

From equation (1), we get the ration of angular frequency and frequency as:

$\frac{\omega }{f}=2\mathrm{\pi }$

But substituting the above equation in equation (2), we get the time period as:

$$\begin{gathered} \omega T=2\mathrm{\pi } \\ T=\frac{2\mathrm{\pi }}{\mathrm{\omega }} \\ =\frac{2\mathrm{\pi }}{40\mathrm{\pi }} \\ =0.05\mathrm{s} \end{gathered}$$

Hence, the value of period of oscillation is 0.05 s

Step 7(e): Calculation of speed of two waves

We have,

$y(x,t)=0.040\left(\sin 5\mathrm{\pi x}\right)\left(\cos 40\mathrm{\pi }t\right)$

From the above superposition displacement formula of the resultant, we can write the two equations as:

$$\begin{gathered} y_1=0.020(\sin 5\mathrm{\pi x})(\cos 40\mathrm{\pi }t) \\ {\mathrm{y}}_2=0.020\sin (5\mathrm{\pi x}+40\mathrm{\pi }t) \end{gathered}$$

Comparing the given equation of the wave with the general equation of the standing wave we get,

From this we have,

$$\begin{gathered} k=5\mathrm{\pi } \\ \mathrm{\omega }=40\mathrm{\pi } \end{gathered}$$

We also have from equation (iii), we get the speed of the waves is given as:

role="math" localid="1661165909881" $$\begin{gathered} V=\frac{40\mathrm{\pi }}{5\mathrm{\pi }} \\ =8.0\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the value of the speed is 8.0 m/s.

Step 8(f): Calculation of amplitude of the two travelling waves

Comparing the given equation of the wave with the general equation of the standing wave we get,

Amplitude of the standing wave,

$y_m=0.040\mathrm{m}$

Amplitude of the two travelling waves in terms of the Amplitude of the standing wave is given as,

role="math" localid="1661166105381" $$\begin{gathered} {\left(y_m\right)}_{s\tan dingwave}=2\times {\left(y_m\right)}_{travellingwave} \\ {\left(y_m\right)}_{travellingwave}=\frac{\left({\left(y_m\right)}_{s\tan dingwave}\right)}{2} \\ =\frac{0.040}{2} \\ =0.020\mathrm{m} \end{gathered}$$

Hence, the value of the amplitude is 0.020 m

Step 9(g): Calculation of time when transverse velocity is zero at all points

The transverse velocity of the wave is given as:

$$\begin{gathered} u=\frac{dy}{dt} \\ =\frac{d(0.040(\sin 5\mathrm{\pi x}\left)\right(\cos 40\mathrm{\pi t}\left)\right)}{dt} \\ =-(0.040)\times \left(40\mathrm{\pi }\right)\times \sin \left(5\mathrm{\pi x}\right)\times \sin \left(40\mathrm{\pi }t\right) \end{gathered}$$

The above function will be zero when$\left(40\mathrm{\pi }t\right)=0$

For the value of thefunction to be zero, the value for$5\mathrm{\pi x}$must be in the range of$0,\mathrm{\pi },2\mathrm{\pi },3\mathrm{\pi }.....$

From this we can say that, for the first time, transverse velocity, u to be zero,

$$\begin{gathered} 40\mathrm{\pi }t=0 \\ t=0\mathrm{s} \end{gathered}$$

Hence, the first time when transverse velocity is zero is 0 s

Step 10(h): Calculation of second time when velocity is zero

For the second time from the given values of phases, transverse velocity to be zero,

$$\begin{gathered} 40\mathrm{\pi }t=\mathrm{\pi } \\ 40t=1 \\ t=\frac{1}{40} \\ =0.025\mathrm{s} \end{gathered}$$

Hence, the second time at which transverse velocity is zero is 0.025 s

Step 11(i): Calculation of the third time when transverse velocity is zero

For the third time from the given ranges, transverse velocity to be zero,

$$\begin{gathered} 40\mathrm{\pi }t=2\mathrm{\pi } \\ 40t=2 \\ t=\frac{2}{40} \\ =0.05\mathrm{s} \end{gathered}$$

Hence, the value of time is 0.05 s