Question

The average power transmitted through a given point on a string supporting a sine wave is $0.40\,watt$ when the amplitude of the wave is $2\,mm$. What average power will be transmitted through this point if its amplitude is increased to $4\,mm$?
A) $0.40\,watt$
B) $0.80\,watt$
C) $1.2\,watt$
D) $1.6\,watt$

Answer 1

Hint: To solve the asked question, you need to use the formula of average power of a wave on a string and see the relation between the power and the amplitude of the wave or the point’s amplitude with the average power. The formula of the average power related the average power of the wave with the density of the string, amplitude of the wave, velocity of the wave and the angular frequency of the wave.

Complete step by step answer:
We will follow the same approach as explained in the hint section of the solution to the asked question. First, we will talk about the formula of the average power of a wave and develop a relation between the average power and the amplitude if all the other factors are kept same, since nothing is told about the other factors like angular frequency of the wave, velocity of the wave, hence, they can be assumed to be the same in both the cases.
First, let’s discuss the formula of the average power of the wave:
${P_{avg}} = \dfrac{1}{2}\mu {\omega ^2}{A^2}V$
Here, $\mu $ is the linear density of the string, which is supposed to be constant in our case,
$\omega $ is the angular frequency of the wave, which is also supposed to be constant in our case,
$V$ is the velocity of the wave, which is also supposed to be constant in our case,
$A$ is the amplitude of the wave, which changes from $2\,mm$ to $4\,mm$ as given in the question and,
${P_{avg}}$ is the average power of the wave.
If consider the linear density of the string $\left( \mu \right)$ , the angular frequency of the wave $\left( \omega \right)$ and the velocity of the wave $\left( V \right)$ as constant, we can see that the average power is directly proportional to the amplitude raised to a power of two, mathematically, this can be represented as:
${P_{avg}} \propto {A^2}$
Let us consider the first case:
The amplitude is given to be ${A_1} = 2\,mm$
The average power is given as ${P_1} = 0.40\,watt$
In the second case, the amplitude has become:
${A_2} = 4\,mm$
The average power is to be found out as:
$\implies$ $\dfrac{{{P_2}}}{{{P_1}}} = \dfrac{{A_2^2}}{{A_1^2}}$
If we substitute the values, we get:
$
\implies {P_2} = {\left( 2 \right)^2} \times 0.40\,watt \\
\implies {P_2} = 1.6\,watt \\
$
Hence, we can see that the option (D) is the correct option as the value matches the value of power that we found out upon solving the question.
$\therefore$ The correct option is D

Note: Many students do not consider the linear density, angular frequency and velocity of the wave as constant, but since no information is given to us about them, we can consider them constant and thus, can easily reach the answer to the question.