Question

Simple Harmonic Motion

Consider a harmonic oscillator with period 0.13 s. If the speed at the equlibrium position is 2.10 m/s, what is the speed of a mass at a point which is displaced by 1.5 cm? (Work this out using conservation of energy).

Please make sure to draw an energy bar chart for this.

Find the answer and show the steps taken to get to the answer.

Evaluate the answer and explain why it makes sense.

Answer 1

According to law of conservation of energy -RV 12 1,1 (21m/s)-(4)-k(1.5× 10-2 m)2 M2 k(1.5×10-2 m) 2 1(2.1 m /s) ½ = Angular frequency is 27T T" M2 の 2π 0.13s -48.33 Hz And, (48.33 Hz) m = (2336) From above equation =y(2 1 m/s)2-4(1 5x10-2 m)2 ½ = 1(2.1 m/s)2-(2336)(1.5×10-2 m)2 M2 =1.97 m/s