Wave graph

[Niha afzal]

My approach for this question is..

The disp time graph will give us the speed .. the speed will give us kinetic energy 0.5mv^2 .

Now the disp time starts with a negative displacement. ..the particle at that point will have a gradient (speed) negative . .. but in the k.e eq the sped gets squared .. therefore the first kinetic energy must be positive having some value.... crossing part B AND D which start from zero.... now how do I chose from A AND C.... if my thoughtprocess is correct ...

[studiot]

Each (air) particle participating in the wave executes simple harmonic motion. That is it is an SHM oscillator.

Note the particle motion is not the wave motion.

 

A particle executing SHM has minumum (zero) velocity at maximum displacement.

So at these points in the SHM cycle, the KE is zero.

 

As you observe, the displacement - time graph starts at maximum negative displacement.

So the KE - time graph must start from zero.

 

Question for you.

 

At what point in the SHM cycle is the velocity (and therefore the KE) a maximum ?

[Niha afzal]

Max k.e when the particle reaches the straight lineee.horizontal axis ...

Therefore in time T it comes to the straight line twice.... .. if negative max is zero velocity and the straight line is max velocity. ... in time T the particle makes 2 curves .. hence the option is D? Is it

Edited March 27, 2016 by Niha afzal

[studiot]

Yes, D. The energy graph has twice the frequency of the displacement graph