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 ...

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 ?

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, 20169 yr by Niha afzal

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