A person is producing a wave in a string by m | vedclass

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

Question

(C) Given frequency $f = \frac{1}{8} \, Hz$.Therefore,time period $T = \frac{1}{f} = 8 \, s$.Given wavelength $\lambda = 24 \, m$.The wave speed $v =

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) Given frequency $f = \frac{1}{8} \, Hz$.Therefore,time period $T = \frac{1}{f} = 8 \, s$.Given wavelength $\lambda = 24 \, m$.The wave speed $v = f \lambda = \frac{1}{8} 	imes 24 = 3 \, m/s$.The time taken by the wave to reach the particle at distance $x = 9 \, m$ is $t_1 = \frac{x}{v} = \frac{9}{3} = 3 \, s$.The particle starts its motion from the mean position (moving up first). To reach the lower extreme for the first time,the particle must complete $\frac{3}{4}$ of its oscillation cycle.The time taken by the particle to reach the lower extreme from the moment the wave reaches it is $t_2 = \frac{3T}{4} = \frac{3}{4} 	imes 8 = 6 \, s$.Total time taken = $t_1 + t_2 = 3 \, s + 6 \, s = 9 \, s$.

Similar Questions

When a wave travels in a medium,the particle displacement is given by $y(x, t) = 0.03 \sin \pi (2t - 0.01x)$,where $y$ and $x$ are in meters and $t$ is in seconds. The phase difference,at a given instant of time,between two particles $25 \ m$ apart in the medium is:

$A$ wave is represented by the equation $y = (0.02 \ m) \sin (5 \pi x - 20 t)$. The minimum distance between two particles always having the same speed is: (Assume $x$ and $t$ are in $SI$ units) (in $m$)

$A$ wave represented by the given equation $Y = A\sin(10\pi x + 15\pi t + \frac{\pi}{3})$,where $x$ is in meters and $t$ is in seconds. The expression represents:

$A$ sine wave of wavelength $\lambda$ is travelling in a medium. The minimum distance between two particles,which always have the same speed,is

Obtain the relation between wave velocity,angular frequency,and angular wave number.

Vedclass Test Series

Exam Paper Generator

Online Exam Module