Question

an object of mass 50g performs SHM of period 6.28s. determine the magnitude of force acting on it when it is 3 cm away from its mean position.​

Answer 1

Answer:

Magnitude of force F = 0.0015 N

Explanation:

Data:

Mass of object = 50 g = 0.05 kg

Time period = T = 6.28 s

Displacement of object from mean position = x = 3 cm = 0.03 m

Required:

Magnitude of force= Force = F = ?

Calculation:

We known that

T = 2π√(M/K)

(T/2π) = √(M/K)

Taking square on both sides we get

T²/4π² = M/K

K = M 4π²/ T²

Putting values we get

K = (0.05) (4π²) / (6.28)² = 0.0500 ≈ 0.05

So K = 0.05 N/m

We known that

According to hooks law

F = -K (x)

Putting values we get

F = -(0.05)(0.03) =   - 0.0015 N

So magnitude of force F = 0.0015 N