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 _____.
Correct Answer: 7
Approach Solution - 1
To solve this problem, we begin by considering the properties of simple harmonic motion. For a particle performing simple harmonic motion with amplitude \( A \), the velocity \( v \) at a displacement \( x \) is given by:
\( v = \omega \sqrt{A^2 - x^2} \)
where \(\omega\) is the angular frequency. At \( x = \frac{2A}{3} \), the initial speed \( v_1 \) is:
\( v_1 = \omega \sqrt{A^2 - \left(\frac{2A}{3}\right)^2} = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \omega \frac{\sqrt{5}A}{3} \)
When the speed triples, the new speed \( v_2 = 3v_1 \):
\( v_2 = 3 \times \omega \frac{\sqrt{5}A}{3} = \omega \sqrt{5}A \)
For the new amplitude \( A' = \frac{nA}{3} \), we write the expression for the new speed:
\( v_2 = \omega \sqrt{A'^2 - x^2} \)
At \( x = \frac{2A}{3} \), \( v_2 \) is also:
\( \omega \sqrt{A'^2 - \left(\frac{2A}{3}\right)^2} = \omega \sqrt{\left(\frac{nA}{3}\right)^2 - \frac{4A^2}{9}} \)
Equating the expressions for \( v_2 \), we have:
\( \omega \sqrt{5}A = \omega \sqrt{\frac{n^2A^2}{9} - \frac{4A^2}{9}} = \omega A \sqrt{\frac{n^2 - 4}{9}} \)
By canceling common terms and solving for \( n \), we get:
\( \sqrt{5} = \frac{\sqrt{n^2 - 4}}{3} \)
\( \sqrt{5} \times 3 = \sqrt{n^2 - 4} \)
Squaring both sides, we obtain:
\( 45 = n^2 - 4 \)
\( n^2 = 49 \)
So, \( n = 7 \) as \( n \) is positive. Thus, the value of \( n \) is 7, which falls within the expected range of 7 to 7.
Approach Solution -2
At \( x = \frac{2A}{3} \),
\[ v = \omega \sqrt{A^2 - \left(\frac{2A}{3}\right)^2} = \omega \sqrt{\frac{5A^2}{9}} = \omega \frac{\sqrt{5}A}{3} \]New amplitude is \( A' \):
\[ v' = 3v = 3 \left( \omega \frac{\sqrt{5}A}{3} \right) = \omega \sqrt{(A')^2 - \left(\frac{2A}{3}\right)^2} \] \[ \omega \sqrt{5}A = \omega \sqrt{(A')^2 - \left(\frac{2A}{3}\right)^2} \]Squaring both sides:
\[ 5A^2 = (A')^2 - \left(\frac{2A}{3}\right)^2 \] \[ (A')^2 = 5A^2 + \left(\frac{4A^2}{9}\right) = \frac{45A^2}{9} + \frac{4A^2}{9} = \frac{49A^2}{9} \] \[ A' = \frac{7A}{3} \] \[ \therefore \, n = 7 \]Top JEE Main Physics Questions
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