Consider a spring-mass simple harmonic oscillator in one dimension. The mass of the particle is m kg and the spring constant is k Nm-1. At a given instant, the extension of the spring is x-meter and the speed of the particle is v ms-1. On the x - v plane, if the graph of v as a function of x is a circle, then the correct option is:
- k = m
- k = 1m
- k = m
- k = m2
Answer
(3) k = m
Solution
By conservation of mechanical energy in SHM:
12 m v2 + 12 k x2 = E
v2(2E/m) + x2(2E/k) = 1
This represents an ellipse in the x - v plane. For this graph to be a circle, the semi-axes must be equal:
2Em = 2Ek k = m
Answer: (3)