A wave in two different strings

[hhegab]

A string has two parts: one with a very high mass density (per unit length), and the other with a very low mass density. A wave with amplitude A moves from the dense part toward the light part. What will be the amplitude of the wave which is transmitted to the light part?

hhegab

 

[himanshu121]

okay what u have done so far on this
ur thoughts

 

[himanshu121]

Some Hints which will initiate you

1) function representing a wave should be bounded
2) function is continuous, Differentiable and bounded

 

[hhegab]

and there should be a node at the meeting poing (between the two strings)

hhegab

 

[himanshu121]

no i don't think so, u have any reasons for that

 

[himanshu121]

consider
$$y_i=A_isin(\omega t-K_1x)$$
$$y_r=A_rsin(\omega t+K_1x)$$
$$y_t=A_tsin(\omega t-K_2x)$$

now function should be continuous
Left Hand Limit = Right Hand Limit (Consider x=0 at the joint)
which give $$A_i+A_r=A_t$$

now function is derivable at x=0
for which u will have $$(A_r-A_i)K_1=A_tK_2$$

solving from two equations u will have
$$A_t=\frac{2K_1A_i}{K_1+K_2}$$
OR
$$A_t=\frac{2\sqrt{\mu_1}}{\sqrt{\mu_1}+\sqrt{\mu_2}}.A_i$$

 

[sridhar_n]

or you can look at the problem like this...

Total Energy is conserved in both the regions of the rope. Thus, consider the wave from the lighter density rope as the incident wave. When this incident wave meets the higher density rope, a part of it gets reflected and a part of it gets transmitted. So, by law of conservation of energy,


Energy of the incident wave = Energy of the reflected wave + Energy of the Transmitted wave.

Also if $$A_{i}$$ is the amplitude of the incident wave,
$$A_{t}$$ is the amplitude of the reflected wave and $$A_{r}$$ is the reflected component of the wave, then,

$$A_{i} = A_{r} + A_{t}$$

Use the total energy equation of the wave,

$$\Delta E = A/2[\mu^2 * A^2 * \Delta x * \sqrt{T/\mu}]$$

where,
The tension $$T$$ is constant throughout the string, $$\mu$$ is the mass per unit length and $$A$$ is the amplitude and $$\Delta x$$ is the displacement which is also assumed to be constant for a small portion of the wave. Thus equating the total energies on both sides, you will get the same answer that himanshu has given you.


Sridhar

 

[himanshu121]

now function is derivable at x=0
for which u will have
$$(A_r-A_i)K_1=A_tK_2$$

Left Hand Derivative = Right Hand Derivative

Also the derivative $$\frac{dy}{dx}$$ represents Strain which will be same at a single point