Question

A particle moving with shm has speed 3cm/s and 4cm/sat displacement 8cm and 6cm from equilibrium posposition find the period of oscilation

Answer 1

Here is your answer ....

Sol: Velocity of a particle during SHM is given by following formula:
                                   v = ωA2-x2where:A = Amplitudex = displacement from mean position(1). As per given that: v = 3 cmsec at x = 8 cmso3 = ωA2-823 = ωA2-64                         .....(1)(2). As per given thatv = 4 cmsec at x = 6 cmso4 = ωA2-624 = ωA2-36                         .....(2)dividing eq ..(2) by (1)169 = A2-36A2-6416A2-16×64 = 9A2-36×97A2 = 1024-324 = 700A = ±10So the value of amplitude will be:A = 10 cm                                        ....(3)now putting value of eq ..(3) in ..(1)3 = ω102-643 = ω100-643 = ω×6ω = 12so the time period will be given as:T = 2πωT = 2π12T = 4π sec.
Hope it helps you ....

Answer 2

Dear Student,

◆ Answer -

T = 12.57 s

● Explanation -

Velocity of particle in SHM is given by formula -

v = w√(A²-x²)

For v = 3 cm/s and x = 8 cm/s,

3 = w√(A²-8²) ...(1)

For v = 4 cm/s and x = 6 cm/s,

4 = w√(A²-6²) ...(2)

Solve (1) and (2),

w = 0.5 rad/s

A = 10 cm

Time period is calculated as -

T = 2π/w

T = 2 × 3.142 / 0.5

T = 12.57 s

Hence, the period of oscilation is 12.57 s.

Thanks dear. Hope this helps you...