A particle in simple harmonic motion starts from it's extreme position of oscillation at time t=0 and has a time period T=6 s. Then the earliest time when it's potential energy equals three times it's kinetic energy will be?

Question

A particle in simple harmonic motion starts from it's extreme position of oscillation at time t=0 and has a time period T=6 s. Then the earliest time when it's potential energy equals three times it's kinetic energy will be? (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Since the SHM starts from extreme position,

x = A cos (ω t)

,
Hence Potential Energy and Kinetic Energies are such that

U \propto cos^{2} (ω t)

and

K E \propto sin^{2} (ω t)

,
Therefore for

U = 3 K E

,

ω t = π /6

\Rightarrow t = T /12

Q. A particle in simple harmonic motion starts from it's extreme position of oscillation at time t=0 and has a time period T=6s. Then the earliest time when it's potential energy equals three times it's kinetic energy will be?

Since the SHM starts from extreme position, x=Acos(ωt), Hence Potential Energy and Kinetic Energies are such that U∝cos2(ωt) and KE∝sin2(ωt), Therefore for U=3KE, ωt=π/6 ⇒t=T/12

Q. A particle in simple harmonic motion starts from it's extreme position of oscillation at time $t = 0$ and has a time period $T = 6 s$ . Then the earliest time when it's potential energy equals three times it's kinetic energy will be?

Since the SHM starts from extreme position, $x = A cos (ω t)$ ,
Hence Potential Energy and Kinetic Energies are such that
$U \propto cos^{2} (ω t)$ and $K E \propto sin^{2} (ω t)$ ,
Therefore for $U = 3 K E$ ,
$ω t = π /6$
$\Rightarrow t = T /12$