Solve Wave Moving w/ Constant Acceleration: Hi Guys!

[Psi-String]

Hi guys, here's the question.
A string, which has linear density \mu, is suspended vertically. Someone produces a wave from the bottom of the string. Please prove that the wave is moving with constant acceleration

My solution:

let x be the distance between a point on the string and the bottom of the string then

T= \mu x g where T is tension

so

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

then

a= \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = \frac{1}{2} g

Am I right or wrong? I'm not sure this really works.
Could someone sovle this problem by other mathematical approach, and tell me why the wave move with constant accleration qualitatively?

Thanks in advanced!

 

[Andrew Mason]

Hi guys, here's the question.
A string, which has linear density \mu, is suspended vertically. Someone produces a wave from the bottom of the string. Please prove that the wave is moving with constant acceleration

My solution:

let x be the distance between a point on the string and the bottom of the string then

T= \mu x g where T is tension

so

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

then

a= \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = \frac{1}{2} g

Am I right or wrong? I'm not sure this really works.
Could someone sovle this problem by other mathematical approach, and tell me why the wave move with constant accleration qualitatively?

The wave analysis breaks down close to the end since the conditions assumed in the analysis (particularly no resistance to changing shape) no longer apply.

But imagine a long heavy cable hanging with its lower end free. Plucking the cable at the middle sends a pulse that travels up the cable at a speed of v = \sqrt{gx} which increases at the rate of g/2 = 4.9m/sec^2 as it travels up the cable. The pulse traveling down the cable starts at the same speed downward but slows at the rate of g/2.

AM

 

[Psi-String]

I see. Thanks!