Power Transmitted Along The String MCQ - Practice Questions with Answers

As a sinusoidal wave moves down a string, the energy associated with one wavelength on the string is transported down the string at the propagation velocity v. From the basic wave relationship the distance traveled in one period is vT = λ, so the energy is transported one wavelength per period of the oscillation.

The energy associated with one wavelength of the wave is:

$$
E_\lambda=\frac{1}{2} \mu \omega^2 A^2 \lambda
$$

so the power transmitted would be :

$$
\begin{aligned}
& P_\lambda=\frac{1}{2} \mu \omega^2 A^2 \frac{\lambda}{T} \\
& \text { since } v=\frac{\lambda}{T}
\end{aligned}
$$

Therefore $P_\lambda=\frac{1}{2} \mu \omega^2 A^2 v$
where $\omega=$ angular frequency,$\mu=$ mass per unit length of string , $A=$ wave amplitude
$v=$ wave propagation velocity

The intensity of the wave-

The flow of energy per unit area of the cross-section of the string in the unit time is known as the intensity of the wave.

$$
\text { As } P=\frac{1}{2} \mu \omega^2 A^2 v
$$

$$
\text { And using } I=\frac{P}{\text { Area }}
$$

we get $I=\frac{\frac{1}{2} \mu \omega^2 A^2 v}{\text { Area }}$
using $\mu=\frac{\text { mass }}{\text { length }}=\frac{m}{l}$ and $V$ Volume $=$ Area $\times$ length
We get $I=\frac{\frac{1}{2} m \omega^2 A^2 v}{\text { length } \times \text { Area }}=\frac{\frac{1}{2} m \omega^2 A^2 v}{\text { Volume }}$
And now using $\rho=\frac{\text { mass }}{\text { volume }}$
we get $I=\frac{1}{2} \rho \omega^2 A^2 v$

Where,

$\rho=$ density
$\omega=$ angular frequency
$A=$ Amplitude
$v=$ Wave speed