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

The phase difference corresponding to a path difference of $x$ is

$A$ wave travelling in the positive $x-$direction with amplitude $A = 0.2\;m$ has a velocity of $v = 360\;m/s.$ If the wavelength $\lambda = 60\;m,$ then the correct expression for the wave is:

The equation of a progressive wave is $y = a \sin(200t - x)$,where $x$ is in meters and $t$ is in seconds. The velocity of the wave is ..... $m/s$.

The equation of a wave is given as $y = 0.07 \sin (12\pi x - 3000\pi t)$. Where $x$ is in $m$ and $t$ is in $s$,then the correct statement is:

Which of the following equations represents a travelling wave?

Vedclass Test Series

Exam Paper Generator

Online Exam Module