# A heavy rope is attached to one end of a lightweight rope [closed]

If one end of a heavy rope is attached to one end of a lightweight rope, a wave can move from the heavy rope into the lighter one.

(a) What happens to the speed of the wave?

(b) What happens to the frequency?

(c) What happens to the wavelength?

My instructor hasn't gone over any of this in class (it's for a reading assignment), so what I've guessed so far just off the equations the book gives.

(a) $v=\sqrt{\frac{F}{\mu }}$

So, as the mass per unit length ($\mu$) goes down, the velocity will increase

(b) $v=\lambda f$

Now I am unsure. There is no way to tell (from this one equation) whether the wavelength ($\lambda$) will increase, decrease, stay constant. Is it determinable at all?

(c) Same problem as with (b).

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## closed as off-topic by John Rennie, Kyle Kanos, ACuriousMind, HDE 226868, Qmechanic♦Aug 19 '15 at 20:33

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, Kyle Kanos, ACuriousMind, HDE 226868, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

Please note that Physics.StackExchange is not a homework help site. Please read this Meta post on asking homework-like questions and this Meta post for "check my work" problems. – Kyle Kanos Aug 19 '15 at 13:27

My intuition is that the frequency should stay the same because the waves in the light rope are caused by the waves in the heavy rope. The point where the ropes attach will oscillate with a common frequency. So, for $(b)$, the frequency would be the same. For $(c)$, use the equation $v= f\lambda$. You already correctly determined that the velocity increases; so, if the frequency stays the same, the wavelength must increase.

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here is an animated version of image

for more in formation refer to http://www.physicsclassroom.com/class/waves/Lesson-3/Boundary-Behavior

and a good tool for playing with