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Two particles (A) and (B) of equal masse...

Two particles (A) and (B) of equal masses are suspended from two massless spring of spring of spring constant `k_(1)` and `k_(2)`, respectively, the ratio of amplitude of (A) and (B) is.

A

(a)`sqrt(k_(1)//k_(2))`

B

(b)`k_(1)//k_(2)`

C

(c)`sqrt(k_(2)//k_(1))`

D

(d)`k_(2)//k_(1)`

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The correct Answer is:
To solve the problem of finding the ratio of amplitudes of two particles (A) and (B) suspended from springs with different spring constants, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the System**: We have two particles A and B of equal mass (let's denote the mass as \( m \)). They are attached to two different springs with spring constants \( k_1 \) and \( k_2 \). 2. **Maximum Velocity in SHM**: The maximum velocity \( v \) of a particle executing simple harmonic motion (SHM) is given by the formula: \[ v = A \cdot \omega \] where \( A \) is the amplitude and \( \omega \) is the angular frequency. 3. **Angular Frequency**: The angular frequency \( \omega \) for a mass-spring system is given by: \[ \omega = \sqrt{\frac{k}{m}} \] where \( k \) is the spring constant and \( m \) is the mass of the particle. 4. **Expressing Maximum Velocity for Both Particles**: For particles A and B, we can express their maximum velocities as: - For particle A: \[ v_1 = A_1 \cdot \omega_1 = A_1 \cdot \sqrt{\frac{k_1}{m}} \] - For particle B: \[ v_2 = A_2 \cdot \omega_2 = A_2 \cdot \sqrt{\frac{k_2}{m}} \] 5. **Setting the Velocities Equal**: Since the problem states that the maximum velocities of both particles are equal, we set \( v_1 = v_2 \): \[ A_1 \cdot \sqrt{\frac{k_1}{m}} = A_2 \cdot \sqrt{\frac{k_2}{m}} \] 6. **Canceling Mass**: Since the masses are equal, we can cancel \( m \) from both sides: \[ A_1 \cdot \sqrt{k_1} = A_2 \cdot \sqrt{k_2} \] 7. **Finding the Ratio of Amplitudes**: Rearranging the equation to find the ratio of amplitudes gives us: \[ \frac{A_1}{A_2} = \frac{\sqrt{k_2}}{\sqrt{k_1}} \] 8. **Final Result**: Thus, the ratio of the amplitudes of particles A and B is: \[ \frac{A_1}{A_2} = \sqrt{\frac{k_2}{k_1}} \] ### Conclusion: The ratio of amplitudes \( A_1 \) and \( A_2 \) is \( \sqrt{\frac{k_2}{k_1}} \).

To solve the problem of finding the ratio of amplitudes of two particles (A) and (B) suspended from springs with different spring constants, we can follow these steps: ### Step-by-Step Solution: 1. **Understand the System**: We have two particles A and B of equal mass (let's denote the mass as \( m \)). They are attached to two different springs with spring constants \( k_1 \) and \( k_2 \). 2. **Maximum Velocity in SHM**: The maximum velocity \( v \) of a particle executing simple harmonic motion (SHM) is given by the formula: \[ ...
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