The particle executing simple harmonic motion has a kinetic energy K₀ cos²(ωt). The maximum values of the potential energy and the total energy are respectively:
(A) K₀ and K₀
(B) 0 and 2K₀
(C) K₀/2 and K₀
(D) K₀ and 2K₀

The correct answer is D. K0K_0 and 2K02K_0.
Explanation:
A particle executing Simple Harmonic Motion (SHM) has its kinetic energy (K.E.) and potential energy (P.E.) varying periodically, while the total energy (T.E.) remains constant.
Kinetic Energy:
The given kinetic energy is:
K=K0cos⁡2(ωt)K = K_0 \cos^2(\omega t)
Here:
• K0K_0 is the maximum kinetic energy of the particle.
• cos⁡2(ωt)\cos^2(\omega t) varies between 0 and 1 during the motion.
Total Energy (T.E.):
In SHM, the total mechanical energy is constant and is the sum of the kinetic and potential energies. At any instant:
T.E.=K+PT.E. = K + P
At the equilibrium position (x=0x = 0):
• All the energy is kinetic: K=K0K = K_0, and P=0P = 0. At the extreme position (x=±Ax = \pm A, where AA is the amplitude):
• All the energy is potential: P=T.E.P = T.E., and K=0K = 0.
Thus, the total energy remains constant and equals 2K02K_0, as derived below.
Maximum Potential Energy (P.E.):
The potential energy in SHM is given by:
P=T.E.−KP = T.E. - K
Substitute K=K0cos⁡2(ωt)K = K_0 \cos^2(\omega t):
P=2K0−K0cos⁡2(ωt)P = 2K_0 - K_0 \cos^2(\omega t) P=K0(2−cos⁡2(ωt))P = K_0 (2 - \cos^2(\omega t))
The maximum value of PP occurs when cos⁡2(ωt)=0\cos^2(\omega t) = 0 (i.e., at the extreme position of the motion):
Pmax=K0(2−0)=K0P_{\text{max}} = K_0 (2 - 0) = K_0
Total Energy:
As derived earlier, the total energy is constant throughout the motion:
T.E.=2K0T.E. = 2K_0
Final Answer:
• The maximum potential energy is K0K_0.
• The total energy is 2K02K_0.
Thus, the correct answer is:
D. K0 and 2K0\boxed{\text{D. } K_0 \text{ and } 2K_0}