Question

The maximum displacement of an oscillatory motion is A=0.49m. Determine the position x at which the kinetic energy of the particle is half it's elastic potential energy

Answer 1

Answer:

The position x, is ± 0.4 m.

Explanation:

The total mechanical energy of the oscillatory motion is given as;

$E_T = U +K.E\\\\E_T = \frac{1}{2} kA^2\\\\ \frac{1}{2} kA^2 = U +K.E\\\\K.E = \frac{1}{2} kA^2 - U ----equation(1)$

When the kinetic energy (E) is half of the elastic potential energy (U);

$K.E = \frac{U}{2} ----equation(2)$

Equate (1) and (2)

$\frac{U}{2} = \frac{1}{2} kA^2 - U\\\\U = kA^2 -2U\\\\U+2U = kA^2\\\\3 U = kA^2\\\\3(\frac{1}{2} kx^2) = kA^2\\\\\frac{3}{2} x^2=A^2\\\\x^2 = \frac{2}{3} A^2\\\\x = \sqrt{\frac{2}{3} A^2} \\\\x = A\sqrt{\frac{2}{3} } \\\\x = 0.49\sqrt{\frac{2}{3} }\\\\x = + /- (0.4 \ m)$

Thus, the position x, is ± 0.4 m.