Tag: from moving to stationary

Questions Related to from moving to stationary

In a standing wave on a string.

  1. In one time period all the particles are simultaneously at rest twice.

  2. All the particles must be at their positive extremes simultaneously once in one time period.

  3. All the particles may be at their positive extremes simultaneously once in a time period.

  4. All the particles are never at rest simultaneously.


Correct Option: A,C
Explanation:

Standing waves are obtained when two waves with same angular frequencies and velocity are superimposed, (They are however moving in the opposite directions).
$x(t) = A\sin(\omega t - kx) + A\sin(\omega t + kx + \delta)$
$x(t) = 2A\cos(kx)\sin(\omega t +\dfrac{\delta}{2})$
For all particles to be simultaneously at rest, the value of the sine function must be equal to zero.
i.e. $\omega t + \dfrac{\delta}{2} = n\pi$
$\Rightarrow$ $t = \dfrac{1}{\omega}(n\pi - \dfrac{\delta}{2})$
$\omega = \dfrac{2\pi}{T}$
$\Rightarrow$ $t = \dfrac{T}{2\pi}(n\pi - \dfrac{\delta}{2})$
$t _{1} =  \dfrac{T}{2\pi}(n\pi - \dfrac{\delta}{2})$
$t _{2} =  \dfrac{T}{2\pi}((n+1)\pi - \dfrac{\delta}{2})$
$t _{2} - t _{1} = \dfrac{T}{2}$


So, the time between this event happening twice is half to time period, so in one cycle this would happen twice. So Option (A)
For the particle to be at positive extreme the sine function can take a value of 1 only.
It can be shown that this happens at an interval of '$T$'
So it will happen twice in a time period if the displacement is max at the start of the time period and once more at the end of the time period,
or else it would happen only once in a time period. Hence Option (C)

Equation of a standing wave is expressed as y = 2A sin$\omega$t coskx. In the equation, quantity $\omega$/k represents

  1. the transverse speed of the particles of the string.

  2. the speed of the component waves

  3. the speed of the standing wave.

  4. a quantity that is independent of the properties of the string.


Correct Option: A
Explanation:

$y=1A\sin\omega t\cos kx\k=\cfrac{2\pi}{\lambda}\ \Rightarrow \cfrac{\omega}{K}=\cfrac{\omega\times \lambda}{2\pi}=\cfrac{\lambda}{2\pi/\omega}=\cfrac{\lambda}{T}=\lambda f=V$

$\Rightarrow\cfrac{\omega}{K}=$ Velocity of wave particles (transverse speed)