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

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

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

Step 4: Equating the Maximum Velocities
We are given that their maximum velocities are the same: v_{max, P} = v_{max, Q}, so: A₁ √(k₁ / m) = A₂ √(k₂ / m).
Simplifying, we get: A₁ √(k₁) = A₂ √(k₂).

Step 5: Finding the Ratio of the Amplitudes
We need to find the ratio of the amplitude of P to the amplitude of Q, which is: A₁ / A₂ = √(k₂ / k₁).
Therefore, the ratio of the amplitudes is: A₁ / A₂ = √(k₂ / k₁).