A musical instrument P produces sound waves o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The energy $E$ of a sound wave is proportional to the square of its amplitude $A$ and the square of its frequency $n$. Mathematically,$E \propto A

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The energy $E$ of a sound wave is proportional to the square of its amplitude $A$ and the square of its frequency $n$. Mathematically,$E \propto A^2 n^2$.Since the energies of the waves produced by $P$ and $Q$ are equal,we have $E_P = E_Q$.This implies $A_P^2 n_P^2 = A_Q^2 n_Q^2$.Given $n_P = n$ and $n_Q = \frac{n}{4}$,we substitute these values into the equation:$A_P^2 n^2 = A_Q^2 (\frac{n}{4})^2$.$A_P^2 n^2 = A_Q^2 (\frac{n^2}{16})$.Dividing both sides by $n^2$,we get $A_P^2 = \frac{A_Q^2}{16}$.$A_Q^2 = 16 A_P^2$.Taking the square root of both sides,we get $A_Q = 4 A_P$.

Similar Questions

The amplitudes of two waves are in the ratio $5 : 2$. If all other conditions for the two waves are the same,what is the ratio of their energy densities?

If the separation between the screen and the source is increased by $2\%$,what would be the effect on the intensity?

$A$ musical instrument $P$ produces sound waves of frequency $n$ and amplitude $A$. Another musical instrument $Q$ produces sound waves of frequency $n/4$. The waves produced by $P$ and $Q$ have equal energies. The amplitude of waves produced by $Q$ will be

Statement-$1$: Two longitudinal waves given by equations $y_1(x, t) = 2a \sin(\omega t - kx)$ and $y_2(x, t) = a \sin(2\omega t - 2kx)$ will have equal intensity.
Statement-$2$: Intensity of waves of given frequency in the same medium is proportional to the square of amplitude only.

Two waves are represented by
$y_1 = 5 \sin 2 \pi(75 t - 0.25 x)$
$y_2 = 10 \sin 2 \pi(150 t - 0.50 x)$
The intensity ratio $\frac{I_1}{I_2}$ of the two waves is

Vedclass Test Series

Exam Paper Generator

Online Exam Module