Problem 14.36
The motion of a particle is given by x(t)=(25cm)cos(15 (rad/s)?t ), where t is in s.
Part A
What is the first time at which the kinetic energy is twice the potential energy?
Given:-
Comparing the given eqn with the general equation as x(t) = A cos(omega*t)
=> A = 25 cm,
K.E. = 2 P.E. ---(1)
But, K.E. = (1/2) kA^2 - (1/2)kx^2 and P.E. = (1/2) kx^2
Substituting above eqns in (1), we get
=> (1/2) kA^2 - (1/2)kx^2 = 2* (1/2) kx^2
=> (1/2) kA^2 = kx^2 + (1/2)kx^2
=> (1/2) kA^2 = (3/2)kx^2 , Cancelling k/2
=> A^2 = 3x^2
=> (25)^2 = 3 * [(25cm) cos(15t)]^2
=> 1/3 = cos2(15t)
=> cos(15t) = sqrt (1/3)
=> 15t = 0.9999
=> t = 0.067 sec.
the motion of a particle is given by x(t)=(25cm)cos(15t) where t is in s
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