Question

A 0.200-kg block on a smooth horizontal surface gains a speed of 28.2 cm/s when it is released from rest at the free end of a spring that is compressed by 3.20 cm. The block is then connected to the free end of the spring to form a mass-spring system.

What is the spring constant? (A) 15.5 N/m; (B) 16.5 N/m; (C) 17.5 N/m; (D) 18.5 N/m; (E) 19.5 N/m.

What is the period of this harmonic oscillator? (A) 0.413 s; (B) 0.513 s; (C) 0.613 s; (D) 0.713 s; (E) 0.813 s.

What happens to the period if the mass of the block is doubled? (A) No change; (B) Doubled; (C) Increased by 40%; (D) Decreased by 40%; (E) Quadrupled.

Answer 1

Given that :

mass of the block, m = 0.2 kg

speed of the block, v = 28.2 cm/s = 0.282 m/s

spring compression distance, x = 3.2 cm = 0.032 m

(a) The spring constant which will be given as :

using conservation of energy, we have

K.E = P.E_spring

(1/2) m v² = (1/2) k x²

m v² = k x² { eq.1 }

inserting the values in above eq.

(0.2 kg) (0.282 m/s)² = k (0.032 m)²

(0.0159 kg.m²/s²) = k (0.001021 m²)

k = (0.0159 kg.m²/s²) / (0.001021 m²)

k = 15.5 N/m

(b) The period of this harmonic oscillator which will be given as :

using an equation, T = 2m / k                                     { eq.2 }

inserting the values in eq.2,

T = 2 (3.14) (0.2 kg) / (15.5 N/m)

T = (6.28) 0.0129 s²

T = (6.28) (0.11357 sec)

T = 0.713 sec

(c) if the mass of the block is doubled, then its period will be increased by 40%