Discussion
Suppose a block of mass 5 kg is attached with the free end of a spring of spring constant 500 N/m. It executes a simple harmonic oscillation on a horizontal frictionless surface if it is displaced to a distance 10 cm from its equilibrium position and released. Then the maximum speed of the block is:
Answer: A
In simple harmonic motion the maximum speed is at the equilibrium position and equals the amplitude times the angular frequency.
\[ \begin{aligned} \omega = \sqrt{\frac{k}{m}} \\ &= \sqrt{\frac{500}{5}} = 10\ \text{rad/s} \end{aligned} \]
\[ v_{max} = A\omega = 0.10 \times 10 = 1\ \text{m/s} \]
Energy gives the same result: \(\tfrac12 kA^2 = \tfrac12 mv^2\) gives \(v^2 = 500 \times 0.01 / 5 = 1\). The amplitude is the 10 cm the block is displaced and released from, which must be changed to metres. The answer is 1 m/s, option A.
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