Question

1. system consists of a massless pulley, a spring of force constant ( k = 10n/m ) and a block of mass ( 0.4kg ). if block is just slightly displaced vertically down from its equilibrium position and released, then the period of vertical oscillations is
2. ( \frac{pi}{2}s ) 2) ( \frac{pi}{5}s ) 3) ( pi s ) 4) ( \frac{2pi}{5}s )

Explanation:

Step1: Determine the equivalent spring constant

When a pulley is used, if the spring constant is \(k\), the equivalent spring constant \(k_{eq}\) for the system is \(4k\). Here \(k = 10N/m\), so \(k_{eq}=4\times10 = 40N/m\).

Step2: Use the formula for the period of a spring - mass system

The formula for the period of a spring - mass system is \(T = 2\pi\sqrt{\frac{m}{k_{eq}}}\). Given \(m = 0.4kg\) and \(k_{eq}=40N/m\).
Substitute the values into the formula: \(T=2\pi\sqrt{\frac{0.4}{40}}\).
Simplify the fraction inside the square root: \(\frac{0.4}{40}=\frac{4}{400}=\frac{1}{100}\).
Then \(T = 2\pi\sqrt{\frac{1}{100}}\).
Since \(\sqrt{\frac{1}{100}}=\frac{1}{10}\), we have \(T=\frac{2\pi}{10}=\frac{\pi}{5}s\).

Answer:

1. \(\frac{\pi}{5}s\)