At the equilibrium position, the cat's potential energy in the spring is zero since the displacement is zero. The elastic potential energy (U) is given by (1/2) k x^2.
The elastic potential energy (U) of a spring in Simple Harmonic Motion (SHM) can be calculated using the formula:
U = (1/2) k x^2
where:
U is the elastic potential energy,
k is the spring constant, and
x is the displacement from the equilibrium position.
In the given scenario, the cat is at its equilibrium position, implying x = 0. At this point, the elastic potential energy is zero, as the spring is neither compressed nor stretched.
The displacement (x) in SHM is given by the amplitude (A) at the equilibrium position, which is 0.050 m. Therefore, the elastic potential energy when the cat is at its equilibrium position is zero:
U = (1/2) k * (0.050)^2 = 0
This result aligns with the understanding that at the equilibrium position, the spring is unstressed, and its elastic potential energy is at a minimum.
The question probable may be:
A thrill-seeking cat with mass 4.00 kg is attached by a harness to an ideal spring of negligible mass and oscillates vertically in SHM. The amplitude is 0.050 m, and at the highest point of the motion the spring has its natural unstretched length. calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring) when the cat is at its equilibrium position. express your answer with the appropriate units.