Question

A particle executes SHM. (a) What fraction of total energy is kinetic and what fraction is potential when displacement is one half of the amplitude? (b) At what value of displacement are the kinetic and potential energies equal?

Answer 1

Fraction of KE of TE = 0.75; Fraction of PE of TE = 0.25; Displacement where kinetic and potential energies are equal is $\frac{A}{\sqrt{2} }$.

Let the equation of the SHM is x = A*sin(wt)

We know, k = mw^2

• Velocity (v) = Aw*cos(wt)
• Kinetic Energy (KE) = 0.5*m*v^2 = 0.5m$(Aw)^{2} cos^{2}(wt)$
• Potential Energy (PE) = 0.5*k*(Asin(wt))^2 = 0.5*m*(wAsin(wt))^2
• Total energy (TE) = 0.5*m*(wA)^2

a)   when displacement is A/2

     From equation of SHM: wt = pi/6

     => t = pi/6w

• KE = 0.5m((Aw)^2)*(3/4)= 0.375*m*(wA)^2
• PE = 0.5*k*(A^2)*(1/4) = 0.125*k*A^2 = 0.125*m*(Aw)^2
• Fraction of KE of TE = 0.75

• Fraction of PE of TE = 0.25

b) KE = PE              => 0.5*m*(wAcos(wt))^2 = 0.5*m*(wAsin(wt))^2

• =>   cos(wt) = sin(wt)
• =>   wt = pi/4
• =>   x = $\frac{A}{\sqrt{2} }$

Answer 2

Thus the values of displacement are equal at 0.707 A

Explanation:

We know that

E  (total)  = 1/2 mω^2 A^2

 K.E. = 1/2 mω^2   (A^2  −x^2  ) and U = 1/2 mω^2  x^ 2

 (a) When x = A/2

K.E. = 1/2 mω^2  3 A  2 /4

K.E./E  total   = 3/4

At, x = A/2, U = 1/2 mω^2   A  2  /4

P.E / E   total   = 1/4

(b) Since K = U  

1/2 mω^2   (A^2   - x^2  ) = 1/2 mω^2 x^ 2

2x^2  = A^2

or x = A/2  = 0.707 A

Thus the values of displacement are equal at 0.707 A