Question

The equation of a wave traveling on a string stretched along the X-axis is given by$y=A{e}^{-{\left(\frac{x}{a}+\frac{t}{T}\right)}^2}$.

(a) Write the dimensions of $A,aandT.$

(b) Find the wave speed.

(c) In which direction is the wave traveling?

(d) Where is the maximum of the pulse located at$t=T?Att=2T?$

Answer 1

Step 1: Given data:

The equation of a wave travelling on a string stretched along the X-axis is given by-

$y=A{e}^{-{\left(\frac{x}{a}+\frac{t}{T}\right)}^2}\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \left(1\right)$

Part (a):

Step 1: Calculate the dimensions of $A,aandT.$

In equation (1)

$y=A{e}^{-{\left(\frac{x}{a}+\frac{t}{T}\right)}^2}$

$A$ is amplitude

Thus the dimension of $A$ is

$\left[A\right]=\left[M^0{L}^1{T}^0\right]$

We know that $y$ is displacement so the dimension of $y$ is also $\left[y\right]=\left[M^0{L}^1{T}^0\right]$

Thus Left-hand side is equal to right-hand side and this can be true only when the term $e^{-{\left(\frac{x}{a}+\frac{t}{T}\right)}^2}$ is dimensionless so this way

and we know that the $xandt$ are distance and time respectively thus the dimension of $\left[x\right]=\left[L\right]and\left[t\right]=\left[T\right]$.

So for the the term $e^{-{\left(\frac{x}{a}+\frac{t}{T}\right)}^2}$to be dimensionless it is required that

$T$ is time period

Thus the dimension of $T$ is $\left[T\right]=\left[M^0{L}^0{T}^1\right]$

The dimension of $a$ is also equal to the dimension of $x$ so the dimension of $a$ is

$\left[a\right]=\left[M^0{L}^1{T}^0\right]$

Therefore the dimension is,

$$\begin{gathered} \left[A\right]=\left[M^0{L}^1{T}^0\right] \\ \left[a\right]=\left[M^0{L}^1{T}^0\right] \\ \left[T\right]=\left[M^0{L}^0{T}^1\right] \end{gathered}$$

Part (b):

Step 2: Calculate the wave speed:

Know that the wave speed is,

$v=\frac{\lambda }{T}\dots \dots \dots \dots \dots \left(2\right)$

Where $\lambda andT$ are wavelength and time period respectively.

Now comparing the general wave equation with the given equation (1)-

The general wave equation

$y=A\sin \left[\left(\frac{t}{T}\right)-\left(\frac{x}{\lambda }\right)\right]$

On comparing we get

$\lambda =a$

Substituting the value of $\lambda$ in equation (2)

Therefore the wave speed is $v=\frac{a}{T}$.

Part (c):

Step 3: Calculating the direction in which the wave is travelling:

Suppose that $y=f\left(t+\frac{x}{v}\right)$ then the wave travels in the negative direction.

If $y=f\left(t-\frac{x}{v}\right)$then the wave travels in the positive direction.

Therefore,

$$\begin{gathered} y=A{e}^{{\left[\left(\frac{x}{a}\right)+\left(\frac{t}{T}\right)\right]}^{-2}} \\ y=A{e}^{-\frac{1}{T}{\left[t+\left(\frac{xT}{a}\right)\right]}^{-2}} \\ y=A{e}^{-\frac{1}{T}\left[t+\frac{x}{v}\right]} \\ y=A{e}^{-f\left[t+\frac{x}{v}\right]} \end{gathered}$$

Since we got $y=f\left(t+\frac{x}{v}\right)$

Hence, the wave is travelling in a negative direction.

Part (d):

Step 4: Calculating the maximum of the pulse located at$t=T?Att=2T?$

The maximum of the pulse is when $y=A$ for this $e^{-{\left(\frac{x}{a}+\frac{t}{T}\right)}^2}$must be equal to $1$.

$e^{-{\left(\frac{x}{a}+\frac{t}{T}\right)}^2}=1$

Using the property rule $e^0=1$

$$\begin{gathered} \left(\frac{x}{a}+\frac{t}{T}\right)=0 \\ \frac{x}{a}=-\frac{t}{T}\dots \dots \dots \dots \dots \dots \left(3\right) \end{gathered}$$

When $t=T$ then substituting the value in equation (3)

$$\begin{gathered} \frac{x}{a}=-\frac{T}{T} \\ x=-a \end{gathered}$$

When $t=2T$ then substituting the value in equation (3)

$$\begin{gathered} \frac{x}{a}=-\frac{2T}{T} \\ x=-2a \end{gathered}$$

Thus, at $t=T$ the maximum pulse $y_{max}=A$ is at $x=-a$that in the negative direction and at $t=2T$ the maximum pulse $y_{max}=A$ is at $x=-2a$that in the negative direction.