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A particle executes SHM of amplitude A and angular frequency omega along a straight line. When it is at a distance (sqrt(3)/2)A from the mean position, an impulsive force instantaneously increases its kinetic energy by (1/2)m*omega²*A². What is the new amplitude of oscillation?

1. sqrt(7)/2 * A
2. sqrt(5)/2 * A
3. 2A
4. sqrt(3) * A

Correct answer: sqrt(7)/2 * A

Solution

In SHM, total energy E = (1/2)m*omega²*A². At x = (sqrt(3)/2)A, PE = (1/2)m*omega²*(3A²/4) = (3/4)E. So old KE = E/4. After impulse, new KE at that point = E/4 + E = 5E/4. New total energy = new KE + PE = 5E/4 + 3E/4 = 2E = m*omega²*A². New amplitude A' satisfies (1/2)m*omega²*A'² = 2E = m*omega²*A², giving A' = sqrt(2)*A... Rechecking: new total = 5E/4 + 3E/4 = 8E/4 = 2E. (1/2)m*omega²*A'² = 2*(1/2)m*omega²*A², so A'² = 2A²... But standard answer for this classic JEE problem is sqrt(7)/2 * A. Let me recheck PE: PE = (1/2)m*omega²*x² = (1/2)m*omega²*(3/4)A² = (3/8)m*omega²*A² = (3/4)*(1/2)m*omega²*A² = (3/4)E. Old KE = E - (3/4)E = E/4 = (1/8)m*omega²*A². Impulse adds (1/2)m*omega²*A² = E. New KE = E/4 + E = 5E/4. New total energy = 5E/4 + 3E/4 = 2E. A'² = 2A², A' = sqrt(2)A. The option sqrt(7)/2*A arises if the added energy is (1/2)m*omega²*A² interpreted differently. Actually for A' = sqrt(7)/2*A: A'² = 7/4 A², so new total = (7/4)*(1/2)m*omega²*A² = 7E/4. This matches if added KE = 7E/4 - E = 3E/4 = (3/8)m*omega²*A², not matching the given. Most reliable: new amplitude = sqrt(2)*A.

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