Question

A particle executes SHM with an amplitude of 10 cm and frequency 2 Hz. At t = 0, the particle is at a point, where potential energy and kinetic energy are same. The equation of displacement of particle is ______.

Options

• 0.1 sin `(4pi"t"+pi/4)`

• 0.1 sin 4πt

• 0.1 cos `(4pi"t"+pi/4)`

• None of these

Answer 1

A particle executes SHM with an amplitude of 10 cm and frequency 2 Hz. At t = 0, the particle is at a point, where potential energy and kinetic energy are same. The equation of displacement of particle is `bbunderline(0.1 sin (4pi"t"+pi/4))`.

Explanation:

Let x = A sin (ωt + Φ)

∴ v = `("d"x)/"dt"` = Aω cos (ωt + Φ)

∴ KE = `1/2"mv"^2=1/2"m""A"^2omegacos^2(omegat +phi)`

∴ (KE)_max = `1/2"mA"^2omega^2`

∴ PE = `1/2"mA"^2omega^2-"KE"`

= `1/2"mA"^2omega^2-1/2"mA"^2omega^2cos^2(omegat+phi)`

= `1/2"mA"^2omega^2sin^2(omega"t"+phi)`

According to problem, KE = PE   (at t = 0)

∴ `1/2"mA"^2omega^2sin^2(omega"t"+phi)`

= `1/2"mA"^2omegacos^2(omega"t"+phi)`

∴ tan² (ωt + Φ) = 1

⇒ tan² (ωt + Φ) = tan² .`pi/4`

⇒ ωt + Φ = `pi/4`

⇒ Φ =  `pi/4`   (∵ t = 0)

∵ x = A sin (ωt + Φ) 

Here, A = 10 cm = 0.1 m

ω = 2πf = 2π × 2 = 4π rad/s

Φ = `pi/4`

∴ x = 0.1 sin  `(4pi"t"+pi/4)`