Question

A spring of force constant 200 N/m is mounted as shown in figure and a mass 2 kg is attached to the free end. If the system is given an initial displacement of 0.05 m and an initial velocity of 2 m/sec, find the amplitude of simple harmonic motion.

Answer 1

\(w = \sqrt{\frac km}\)

\(=\sqrt{\frac{200}2} \)

\(= 10\)

\(v = w\sqrt{A^2 - y^2}\)

\(2 = 10 \sqrt{A^2 - (0.05)^2}\)

⇒ \(0.2 = \sqrt{A^2 - (0.05)^2}\)

⇒ \(A = \sqrt{(0.2)^2 + (0.05)^2}\)

\(= 0.206m\)