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. About the author Maya
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 , [tex]x = A \ sin (wt + \alpha ) \ ; \ v = Aw \ cos(wt+\alpha )[/tex] 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 Reply
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 ,
[tex]x = A \ sin (wt + \alpha ) \ ; \ v = Aw \ cos(wt+\alpha )[/tex]
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