If the frequency of a wave is increased by 25 | vedclass

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

Question

(B) The speed of a wave $v$ is given by $v = f \lambda$. Since the medium is not changed,the speed $v$ remains constant.Therefore,$f_1 \lambda_1 = f_2

Vedclass · Accepted Answer

$20 \%$ decrease

Vedclass · Answer

(B) The speed of a wave $v$ is given by $v = f \lambda$. Since the medium is not changed,the speed $v$ remains constant.Therefore,$f_1 \lambda_1 = f_2 \lambda_2$,which implies $f \propto \frac{1}{\lambda}$.Given that the frequency is increased by $25 \%$,the new frequency $f_2 = f_1 + 0.25 f_1 = 1.25 f_1 = \frac{5}{4} f_1$.Using the inverse relationship,$\frac{\lambda_2}{\lambda_1} = \frac{f_1}{f_2} = \frac{f_1}{1.25 f_1} = \frac{1}{1.25} = \frac{4}{5} = 0.8$.This means $\lambda_2 = 0.8 \lambda_1$.The percentage change in wavelength is $\frac{\lambda_2 - \lambda_1}{\lambda_1} 	imes 100 = \frac{0.8 \lambda_1 - \lambda_1}{\lambda_1} 	imes 100 = -0.2 	imes 100 = -20 \%$.The negative sign indicates a $20 \%$ decrease in wavelength.

If the frequency of a wave is increased by $25 \%$,then the change in its wavelength is (medium not changed).

Similar Questions

Two points are located at a distance of $10\; m$ and $15 \;m$ from the source of oscillation. The period of oscillation is $0.05 \;s$ and the velocity of the wave is $300 \;m/s$. What is the phase difference between the oscillations of the two points?

Write the definition of wave speed and derive $v = \frac{\omega}{k}$.

For a transverse wave travelling along a straight line,the distance between two peaks (crests) is $5 \, m$,while the distance between one crest and one trough is $1.5 \, m$. The possible wavelengths (in $m$) of the waves are

The figure represents the instantaneous picture of a transverse harmonic wave traveling along the negative $x$-axis. Choose the correct alternative$(s)$ related to the movement of the nine points shown in the figure. The stationary point$(s)$ is/are:

$A$ transverse wave propagates in a medium with a velocity of $1450 \, m/s$. The distance between the nearest points at which the oscillations of the particles are in the opposite phase (phase difference of $\pi$) is $0.1 \, m$. What is the frequency of the wave in $Hz$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module