Final answer:
The first time at which the kinetic energy is twice the potential energy is t = 0 or t = 0.5s.
Step-by-step explanation:
In order to find the first time at which the kinetic energy is twice the potential energy, we need to equate the two energies and solve for t. The kinetic energy (K) is given by: K = (1/2)m(v^2), where m is the mass of the particle and v is its velocity. The potential energy (U) is given by: U = (1/2)k(x^2), where k is the spring constant and x is the displacement of the particle.
In this case, we have x(t) = (25cm)cos(15t). To find the velocity, we take the derivative of x(t) with respect to t: v(t) = -(25cm)(15)sin(15t). Now we can calculate the kinetic and potential energy at time t. Equating both energies and solving for t, we get:
K(t) = U(t)(1/2)m(v^2) = (1/2)k(x^2)(1/2)m((-(25cm)(15)sin(15t))^2) = (1/2)(k)((25cm)cos(15t))^2t = 0 or t = 0.5s