Question

A particle performs simple harmonic motion with amplitude \( A \). Its speed is increased to three times at an instant when its displacement is \( \frac{2A}{3} \). The new amplitude of motion is \( \frac{nA}{3} \). The value of \( n \) is _____.

Answer 1

Correct Answer: 7

This problem is addressed by first examining simple harmonic motion properties. For a particle undergoing simple harmonic motion with amplitude \( A \), its velocity \( v \) at a displacement \( x \) is expressed as: \( v = \omega \sqrt{A^2 - x^2} \), where \(\omega\) denotes the angular frequency. At a displacement of \( x = \frac{2A}{3} \), the initial speed, denoted as \( v_1 \), is calculated as: \( v_1 = \omega \sqrt{A^2 - \left(\frac{2A}{3}\right)^2} = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \omega \frac{\sqrt{5}A}{3} \). When the speed is tripled, the new speed, \( v_2 \), becomes \( 3v_1 \): \( v_2 = 3 \times \omega \frac{\sqrt{5}A}{3} = \omega \sqrt{5}A \). For a new amplitude \( A' = \frac{nA}{3} \), the expression for the new speed is: \( v_2 = \omega \sqrt{A'^2 - x^2} \). At \( x = \frac{2A}{3} \), \( v_2 \) is also given by: \( \omega \sqrt{A'^2 - \left(\frac{2A}{3}\right)^2} = \omega \sqrt{\left(\frac{nA}{3}\right)^2 - \frac{4A^2}{9}} \). Equating the expressions for \( v_2 \) yields: \( \omega \sqrt{5}A = \omega \sqrt{\frac{n^2A^2}{9} - \frac{4A^2}{9}} = \omega A \sqrt{\frac{n^2 - 4}{9}} \). After canceling common terms and solving for \( n \), we arrive at: \( \sqrt{5} = \frac{\sqrt{n^2 - 4}}{3} \) and \( \sqrt{5} \times 3 = \sqrt{n^2 - 4} \). Squaring both sides results in: \( 45 = n^2 - 4 \), which simplifies to \( n^2 = 49 \). Therefore, \( n = 7 \) since \( n \) must be positive. The value of \( n \) is determined to be 7.