Why vthe frequency of oscillation of potential energy and kinetic energy is twice as that of displacement or velocity or acceleration of a particle executing s.H.M.
Answers
Answer:
First let me help you to visualise it, and then do the maths.
We take a simple pendulum as an example.
We assume zero friction loss and zero air resistance. Further we assume that the string is light and inextensible. These assumptions (especially the ones about the string) may seem trivial but actually are important assumptions.
At either of the extreme positions of the pendulum, the Kinetic Energy (KE) = 0 while the Potential Energy (PE) = maximum. Let us call the extreme positions as E1 and E2 and the centre position as C.
Assuming for the sake of the argument that the pendulum bob is taken to position e1 and released with zero initial speed. Under gravity, the pendulum moves to C, then to E2, then back to C, then again E1, then again C and so on......
The above E1-C-E2-C-E1 is one cycle.
As you can see, the KE goes from zero (at E1) to max (at C) then again to zero (at E2) then again to max (at C).
[Conversely, the PE goes from maximum (at E1) to minimum (at C) then again to maximum (at E2)]
The above is one cycle of the pendulum SHM but as you can see, the KE goes through two cycles in this time. How? Let us take "KE E1" as KE at E1 and "KE E2" as KE at E2 and "KE C" as KE at C = KE Max.
Obviously, "KE E2" = "KE E1" = 0. Also "KE C" = KE Max.
Thus in the "E1-C-E2-C-E1" cycle the KE goes through two cycles viz
"KE E1" - "KE C"- "KE E2"
&
"KE E2" - "KE C"- "KE E1"
since both are equivalent to 0-KE Max-0
So if the pendulum frequency is f, the KE frequency is 2f.
The maths is quite simple too.
For the example given above, at time t =0, the pendulum bob is at the max displacement from the centre position C. So we can take the equation for position (displacement) at time t as x(t) = ACos(wt) where w = angular frequency and A is Amplitude i.e. max displacement from C.
[In the general SHM equation, there is also a phase term, generally written as phi but we can take it as zero here because of the way we have defined the initial conditions. In any case, the phase does not change the "2f" result we will get]
[The general SHM equation can be written using a Sine term also. Here we have taken a Cosine term since it makes sense because of the way we have defined the initial conditions i.e. x(t) at t=0 is A. If we put a Sine term, then given this initial condition we will have to write x(t) = A(Sin(wt+π/2))]
To get the speed at time t (let us call it v(t)), we can differentiate x(t) with respect to time t. This gives us v(t) = -wASin(wt).
KE = (1/2)m(v^2) where m is the mass (here of the pendulum bob)
Thus KE = (1/2)(m)((w^2)(A^2)(Sinwt)^2)
We know that w=2πf
So KE = (1/2)(m)((w^2)(A^2)(Sin2πft)^2 where only the (Sin2πft)^2 term is time dependent and thus will determine frequency. You can visualise the (Sin2πft)^2 curve in your mind and see that it has frequency 2f because of the square term (^2)
More rigorously-
Basic trigonometry teaches us that (SinP)^2 = (1/2)(1-Cos2P)
Using this, we get KE = (1/2)(m)((w^2)(A^2)((1/2)(1-Cos4πft).
Here only the Cos 4πft term is time dependent and thus will determine frequency. It is evident that this term has a frequency 2f.
Incidentally, and on a similar basis, you can also visualise and/or prove that the PE has frequency 2f.