Question

Two identical point masses P and Q, suspended from two separate massless springs of spring constants \(k_1\) and \(k_2\), respectively, oscillate vertically. If their maximum velocities are the same, the ratio of the amplitude of P to the amplitude of Q is :

• A. \( \sqrt{\frac{k_2}{k_1}} \)
• B. \( \sqrt{\frac{k_1}{k_2}} \)
• C. \( \frac{k_2}{k_1} \)
• D. \( \frac{k_1}{k_2} \)

Hint:

The maximum velocity in SHM is \( v_{max} = A \omega = A \sqrt{k/m} \). Equate the maximum velocities for the two masses and solve for the ratio of their amplitudes.

Answer 1

The Correct Option is A

Step 1: Angular Frequency and Maximum Velocity Formulas
For simple harmonic motion (SHM), angular frequency \( \omega \) is \( \omega = √(k / m) \), where \( k \) is the spring constant and \( m \) is the mass.
The velocity \( v(t) \) of a particle in SHM is \( v(t) = -ωA sin(ωt + φ) \), where \( A \) is the amplitude and \( φ \) is the phase constant.
The maximum velocity \( v_{max} \) is given by \( v_{max} = ωA \).

Step 2: Calculations for Mass P
For mass P, the angular frequency is \( \omega₁ = √(k₁ / m) \) and the amplitude is \( A₁ \).
Its maximum velocity is \( v_{max, P} = ω₁ A₁ = A₁ √(k₁ / m) \).

Step 3: Calculations for Mass Q
For mass Q, the angular frequency is \( \omega₂ = √(k₂ / m) \) and the amplitude is \( A₂ \).
Its maximum velocity is \( v_{max, Q} = ω₂ A₂ = A₂ √(k₂ / m) \).

Step 4: Equating Maximum Velocities
Given that their maximum velocities are equal: \( v_{max, P} = v_{max, Q} \).
Thus, \( A₁ √(k₁ / m) = A₂ √(k₂ / m) \).
Simplifying yields \( A₁ √(k₁) = A₂ √(k₂) \).

Step 5: Determining the Ratio of Amplitudes
The ratio of the amplitude of P to the amplitude of Q is \( A₁ / A₂ \).
From the previous step, \( A₁ / A₂ = √(k₂ / k₁) \).
Therefore, the ratio of the amplitudes is \( A₁ / A₂ = √(k₂ / k₁) \).