Question

The period of vibration of a body of mass 25gm attached to a spring, vibrating on a smooth horizontal surface, when it is displaced 10cm to the right of its extreme position, the period of vibration is 1.57sec and the velocity at the end of this displacement is 0.4m/s. Determine the i) Spring constant ii) Total energy iii) Amplitude

Answer 1

Given :

Mass of the body = 25 g = 0.025 kg

Time period = 1.57 s

Displacement = 10 cm = 0.1 m

Velocity = 0.4 m/s

To Find :

• Spring constant
• Total energy
• Amplitude

Solution :

❖ First of all we have to find angular velocity of the body$.$

➙ ω = 2π / T

Where T denotes time period

By substituting the given values;

➙ ω = (2 × 3.14) / 1.57

➙ ω = 4 rad s‾¹

A] Spring constant :

➠ k = m ω²

➠ k = 0.025 × 4

➠ k = 0.1 kg s‾²

B] Amplitude of body :

Velocity of body at a distance of x from the mean position is given by

• v = ω (A - x)

➠ 0.4 = 4 (A - 0.1)

➠ 0.1 = A - 0.1

➠ A = 0.2 m

C] Total energy :

➠ E = 1/2 kA²

➠ E = 1/2 × 0.1 × (0.2)²

➠ E = 0.004/2

➠ E = 0.002 J = 2 mJ

Answer 2

Answer:

Given :

Mass of the body = 25 g = 0.025 kg

Time period = 1.57 s

Displacement = 10 cm = 0.1 m

Velocity = 0.4 m/s

To Find :

Spring constant

Total energy

Amplitude

Solution :

❖ First of all we have to find angular velocity of the body..

➙ ω = 2π / T

Where T denotes time period

By substituting the given values;

➙ ω = (2 × 3.14) / 1.57

➙ ω = 4 rad s‾¹

A] Spring constant :

➠ k = m ω²

➠ k = 0.025 × 4

➠ k = 0.1 kg s‾

B] Amplitude of body :

Velocity of body at a distance of x from the mean position is given by

v = ω (A - x)

➠ 0.4 = 4 (A - 0.1)

➠ 0.1 = A - 0.1

➠ A = 0.2 m

C] Total energy :

➠ E = 1/2 kA²

➠ E = 1/2 × 0.1 × (0.2)²

➠ E = 0.004/2

➠ E = 0.002 J = 2 mJ