Question

A spring mass system is placed on smooth horizontal table. From equilibrium position the mass is pulled rightwards to point A and released from rest. The object passes equilibrium position and moves until it stops momentarily at point B. Total time taken to move from A to B is 2 seconds. Mass of the object is 100 grams and total energy of the system is 0.9 J. What is the frequency of vibration and distance between A and B.

Answer 1

T½ = T/2 = 2 sec

T = 4 sec

Frequency = 1/T = 1/4

= 0.25 / sec
A at excel eeeel HA 2A

Total mechanical energy = 0.9 J

1/2 mw²A² = 0.9

{ m = 0.1 kg , w = 2π/T = 2π/4 = π/2 }

1/2 × 0.1 × (π/2)² A² = 0.9

π² A² / 4 = 18

π A = √ 4×18

π A = 2 × 3 × 1.4

A = 6 × 1.4 / 3.14

A = 2.675 m

So the distance travelled between A and B

= 2 × A

= 5.35 m