Question

The free lower end of the rope is struck sharply at time t=0. What is the time t it takes the resulting wave on the rope to travel to the ceiling, be reflected, and return to the lower end of the rope? Express your answer in terms of L and constants such as g (the magnitude of the acceleration due to gravity), π, etc.

Answer 1

Answer:

$t=2\sqrt{\frac{L}{g}}$

Explanation:

The equation of the velocity of wave is given by:

$v=\sqrt{\frac{T}{\mu}}$

Here:

• μ is the mass per length unit (μ=m/L).

• T is the tension in each point of the rope with a specific velocity. (T=mg=μyg)

So the velocity will be:

$v=\sqrt{\frac{\mu yg}{\mu}}=\sqrt{yg}$

We know that the velocity is the variation of the position with respect to time so:

$v=\frac{dy}{dt}=\sqrt{yg}$

$\frac{dy}{\sqrt{yg}}=dt$

Let's take the integral in both sides:

$\int^{t}_{0}dt=\int^{L}_{0}\frac{dy}{\sqrt{yg}}$

Solving these two integrals we have:

$t=2\sqrt{\frac{y}{g}}|^{L}_{0}=2\sqrt{\frac{L}{g}}$

Therefore the time will be $2\sqrt{\frac{L}{g}}$

I hope it helps you!

           

Answer 2

Velocity of the wave can be defined as the distance traveled by the wave in a given period of time.

From the equation of velocity of wave:

$v &=\sqrt{\dfrac{T}{\mu}}$

where,

$\mu$ = mass per length

T = tension at each point of the rope with a velocity.

The values substituted in the above equation are given as:

$v = \sqrt{\dfrac{\mu \times y\times g}{\mu}}\\v = \sqrt{y \times g}$

Now, integrating both the sides, we get:

$\int\limits^t_0 {dt} &= \int\limits^L_0 {\dfrac{dy}{\sqrt{y g}}}$

Solving the integration, we get:

$t = \left(2\sqrt{\dfrac{y}{g}}\right)^L_0 &= 2\sqrt{\dfrac{L}{g}}$

Thus, the time will be given as $2 \sqrt{\dfrac{L}{g}}$.

To know more about wave velocity, refer to the following link:

https://brainly.com/question/1292129