Question

A 1.40 kg block is attached to a spring with spring constant 14 N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 47 cm/s .

Part A

What is the amplitude of the subsequent oscillations?

Express your answer in centimeters.

Part B

What is the block's speed at the point where x=0.65A?

Express your answer in centimeters per second.

Answer 1

Solution:-

Given –

Mass of the block-1.40kg

Spring constant = 14N/m

Speed – 47cm/s = 0.47m/s

Part A)      Calculate amplitude of subsequent oscillation

Kinetic energy (KE) = Potential energy (PE)

½*mv² = ½*KA²

½*(1.40)*(0.47)² = ½*(14)*(A) ²

Both side ½ get cancelled,

(1.40)*(0.47)² = (14)*(A) ²

0.30926 = 14A²

A = 0.1486m

A = 14.8 cm

Amplitude of subsequent oscillation is A = 0.1486m or A = 14.8cm

Part B)     Calculate block speed

X = 0.65A

V = (2π/T)*√ (A²-X²)

T = 2π√m/k

T = 1.98s

V = (2π/1.98)*√ (A²-X²)

X = 0.65A

V = (2π/1.98)*√[(0.1486)²-(0.65A)²]

V =0.3586 m/s

V = 35.8 cm/s

Speed of the block is = V = 35.8 cm/s