Question

# A 200-g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250 s. The total energy of the system is 2.00 J. Find (a) the force constant of the spring and (b) the amplitude of the motion.

Solution

## We are given $$m= 200 \ g, T= 0.250 \ s, E= 2.00 \ J,$$ and                      $$\omega= \dfrac{2 \pi}{T} = \dfrac{2 \pi}{0.250 } = 25.1 \ rad/s$$(a) $$k= m \omega^2 = (0.200 \ kg)(25.1 \ rad/s)^2 = \boxed{126 \ N/m}$$(b) $$E= \dfrac{kA^2}{2} \implies A = \sqrt{\dfrac{2E}{k}}= \sqrt{\dfrac{2(2.00 \ J)}{126 \ N/m}}= \boxed{0.178 \ m}$$Physics

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