Question

Energy is transmitted at rate${\mathit{P}}_{\mathbf{1}}$by a wave of frequency ${\mathit{f}}_{\mathbf{1}}$ on a string under tension ${\mathit{\tau }}_{\mathbf{1}}$. What is the new energy transmission rate ${\mathit{P}}_{\mathbf{2}}$ in terms of${\mathit{P}}_{\mathbf{1}}$(a) if the tension is increased to${\mathit{\tau }}_{\mathbf{2}}\mathbf{=}\mathbf{4}{\mathit{\tau }}_{\mathbf{1}}$ and (b) if, instead, the frequency is decreased to${\mathit{f}}_{\mathbf{2}}\mathbf{=}{\mathit{f}}_{\mathbf{1}}\mathbf{/}\mathbf{2}$?

Short answer

a) New energy transmission $P_2$in terms of $P_1$if the tension is increased to ${\tau }_2=4{\tau }_1$is$2P_1$.

b) New energy transmission $P_2$in terms of $P_1$if the frequency is increased to$f_2=f_1/2$ is$P_1/4$

Step-by-step solution

Given data

Energy is transferred at rate $P_1$with frequency $f_1$under tension ${\tau }_1$ of the string.

Understanding the concept of the power of waves

We use the concept of the power of waves which involves linear mass density, velocity, angular velocity, and amplitude. Using this equation, we can find the relation between power and tension, and frequency. Then, we can find the new power in terms of initial power by taking the ratio.

Formulae:

The power rate of a body in motion,$P=\frac{1}{2}\mu v{\omega }^2{y}_m^2..............\left(1\right)$

The velocity of the string, $v=\sqrt{\frac{\tau }{\mu }}.............\left(2\right)$

Step 3(a): Calculation of power if tension is increased

If the tension is increased to${\tau }_2=4{\tau }_1$

Plugging the values of velocity from equation (2) in equation (1), we get the power as:

$$\begin{gathered} P=\frac{1}{2}\mu \sqrt{\frac{\tau }{\mu }}{\left(2\mathrm{\pi f}\right)}^2{y}_m^2 \\ =\frac{1}{2}\mu \sqrt{\tau }4{\mathrm{\pi }}^2{\mathrm{f}}^2{\mathrm{y}}_{\mathrm{m}}^2 \end{gathered}$$

Here,andare the variables and the remaining terms are constant terms, so the power is directly proportional to the tension and frequency. We can write, the power of two waves as given:

$P_1=\frac{1}{2}\sqrt{\mu }\sqrt{{\tau }_1}4{\mathrm{\pi }}^2{\mathrm{f}}^2{\mathrm{y}}_{\mathrm{m}}^2...............\left(3\right)$

Since,, we can write,

role="math" localid="1660983254149" $$\begin{gathered} P_2=\frac{1}{2}\sqrt{\mu }\sqrt{{\tau }_2}4{\mathrm{\pi }}^2{\mathrm{f}}^2{\mathrm{y}}_{\mathrm{m}}^2 \\ =2\left[\frac{1}{2}\sqrt{\mathrm{\mu }}\sqrt{{\mathrm{\tau }}_1}4{\mathrm{\pi }}^2{\mathrm{f}}^2{\mathrm{y}}_{\mathrm{m}}^2\right] \\ =2{\mathrm{P}}_1\left(\mathrm{from}\mathrm{equation}\left(1\right)\right) \end{gathered}$$

Hence, the required value of power is$2P_1$

Step 4(b): Calculation of power if frequency is decreased

New energy transmission $P_2$in terms of $P_1$if the frequency is decreased to $f_2=\frac{f_1}{2}$:

If the tension is the same but frequency is halved, we can write the power of two waves as given:

$$\begin{gathered} P_1=\frac{1}{2}\sqrt{\mu }\sqrt{\tau }4{\mathrm{\pi }}^2{\mathrm{f}}_1^2{\mathrm{y}}_{\mathrm{m}}^2...............\left(4\right) \\ {P}_2=\frac{1}{2}\sqrt{\mu }\sqrt{\tau }4{\mathrm{\pi }}^2{\mathrm{f}}_2^2{\mathrm{y}}_{\mathrm{m}}^2 \\ =\frac{1}{2}\sqrt{\mathrm{\mu }}\sqrt{\mathrm{\tau }}{\left(\frac{{\mathrm{f}}_1}{2}\right)}^{2}4{\mathrm{\pi }}^2{\mathrm{f}}_{}^2{\mathrm{y}}_{\mathrm{m}}^2 \\ =\frac{1}{4}\left[\frac{1}{2}\sqrt{\mathrm{\mu }}\sqrt{\mathrm{\tau }}4{\mathrm{\pi }}^2{\mathrm{f}}_1^2{\mathrm{y}}_{\mathrm{m}}^2\right] \\ =\frac{{\mathrm{P}}_1}{4} \end{gathered}$$

Hence, the required value of power is$\frac{P_1}{4}$