Question

A 4.3 kg block hangs from a spring with constant 2120 N/m the block is pulled down 6.20 cm from the equilibrium position and given an initial velocity of 1.5 m/s back towards equilibrium what is frequency of the motion and what is total mechanical energy of the motion​

Answer 1

Answer :

Spring constant = 2120 N/m

Amplitude of motion = 6.20 cm

Maximum velocity of block = 1.5 m/s

We have to find frequency of the motion and total mechanical energy of the motion$.$

★ Maximum velocity of particle executing SHM is given by

• v = A ω

v denotes velocity

A denotes amplitude

ω denotes angular frequency

• v = 1.5 m/s
• A = 6.20 cm = 0.062 m

By substituting the values, we get

➝ v = A ω

➝ 1.5 = 0.062 ω

➝ ω = 24.2 rad/s

We know that, ω = 2π × f

➝ 24.2 = 2 × 3.14 × f

➝ f = 24.2/6.28

➝ f = 3.85 Hz _____ (A)

Total mechanical energy of particle executing SHM at any point is given by

➝ E = 1/2 k A²

• k denotes spring constant
• A denotes amplitude

➝ E = 1/2 × 2120 × (0.062)²

➝ E = 1/2 × 8.14

➝ E = 4.07 J ______ (B)