The number of waves counted by an observer on | vedclass

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

Question

(A) Given:Number of waves $(n)$ = $54$Time $(t)$ = $1 	ext{ minute} = 60 	ext{ seconds}$Wavelength $(\lambda)$ = $5 	ext{ m}$Step $1$: Calculate th

Vedclass · Accepted Answer

$4.5$

Vedclass · Answer

(A) Given:Number of waves $(n)$ = $54$Time $(t)$ = $1 	ext{ minute} = 60 	ext{ seconds}$Wavelength $(\lambda)$ = $5 	ext{ m}$Step $1$: Calculate the frequency $(f)$ of the waves.Frequency is the number of waves per second.$f = \frac{n}{t} = \frac{54}{60} 	ext{ Hz} = 0.9 	ext{ Hz}$Step $2$: Calculate the velocity $(v)$ of the waves.The relationship between velocity, frequency, and wavelength is given by:$v = f 	imes \lambda$$v = 0.9 	ext{ Hz} 	imes 5 	ext{ m} = 4.5 	ext{ m/s}$Therefore, the velocity of the waves is $4.5 	ext{ m/s}$.

The number of waves counted by an observer on the sea-coast in one minute is $54$. If the wavelength of the waves is $5 \text{ m}$, then the velocity of the waves in $\text{m/s}$ will be:

Similar Questions

$A$ 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:

Velocity of sound waves in air is $V \ m/s$. For a particular sound wave in air,a path difference of $x \ cm$ is equivalent to a phase difference of $n\pi$. The frequency of this wave is:

$A$ simple harmonic progressive wave is represented as $y = 0.03 \sin \pi (2 t - 0.01 x) \ m$. At a given instant of time,the phase difference between two particles $25 \ m$ apart is

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

The equation of a wave is given by $y = 10 \sin \left( \frac{2 \pi}{45} t + \alpha \right)$. If the displacement is $5 \text{ cm}$ at $t = 0$,then the total phase at $t = 7.5 \text{ s}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module