Question

A particle of mass 200 g performs S.H.M. of amplitude 0.1 m and period 3.14 seconds. Find its K.E. when it is

at a distance of 0.03 m from the mean position.​

Answer 1

Given : - particel is in simple harmonic motion (S.H.M)

Mass (m) = 200 gm or 0.2 kg

Amplitude (A) = 0.1 m

Time period (T) = 3.14 seconds or π seconds

Need to find : -  Kinetic energy (K.E) when particel is at a distance, x = 0.03 m form the mean position

Theroy : -

we know for a particle in SHM its postion and velocity at any time t can be given by ,

$x = A \ sin (wt + \alpha ) \ ; \ v = Aw \ cos(wt+\alpha )$

we know Kinetic energy (K.E) =  1/2 mv²

⇒ KE = 1/2 m ω²A²cos²(ωt + ∝)   ; at ∝ = 0

⇒ KE = 1/2 k A² ( 1 - sin²(ωt) )  

 ∴ KE = 1/2 mω² ( A²- x² )

Solution  : -  

we know , ω = 2π / T = 2π / π

⇒ ω = 2 sec ⁻¹

Now, KE = 1/2 mω² ( A²- x² )

⇒ KE = 1/2 x 0.2 x 2² x ( 0.1² - 0.03² )

∴ KE = 3.64 x 10⁻³ joules