A mass 0.9 kg, attached a horizontal spring, executes SHM with an amplitude A1. When this mass passes through its mean position, then a smaller mass of 124 g is placed over it and both masses move together with amplitude A2. If the ratio A1/A2 is α/(α – 1), then the value of α will be _____.
Correct Answer: 16
Approach Solution - 1
To find the value of \( \alpha \), let's analyze the situation step-by-step based on the principles of physics related to Simple Harmonic Motion (SHM) and conservation of momentum.
1. Initially, we have a mass \( m_1 = 0.9 \, \text{kg} \) executing SHM with amplitude \( A_1 \).
2. As it passes through the mean position, a smaller mass \( m_2 = 0.124 \, \text{kg} \) is added. The new total mass is \( m_1 + m_2 \).
3. By conservation of momentum at the mean position:
\[ m_1 \cdot v_1 = (m_1 + m_2) \cdot v_2 \] where \( v_1 \) and \( v_2 \) are the velocities of masses at the mean position before and after adding the smaller mass, respectively.
4. In SHM, velocity at the mean position is maximum and given by \( v = A \cdot \omega \), where \( \omega \) is the angular frequency. Thus, \( \omega_1 = \sqrt{\frac{k}{m_1}} \) and \( \omega_2 = \sqrt{\frac{k}{m_1 + m_2}} \).
5. Using conservation of momentum and the relation \( A_1 \omega_1 = A_2 \omega_2 \):
\[ A_1 \sqrt{\frac{k}{m_1}} = A_2 \sqrt{\frac{k}{m_1 + m_2}} \]
6. By simplifying, we get:
\[ \frac{A_1}{A_2} = \sqrt{\frac{m_1 + m_2}{m_1}} \]
7. Given \( \frac{A_1}{A_2} = \frac{\alpha}{\alpha - 1} \), equate both expressions:
\[ \frac{\alpha}{\alpha - 1} = \sqrt{\frac{0.9 + 0.124}{0.9}} \]
8. Calculate the right side:
\[ \sqrt{\frac{1.024}{0.9}} \approx \sqrt{1.1378} \approx 1.066 \]
9. Solving \( \frac{\alpha}{\alpha - 1} = 1.066 \), multiplying each side by \( \alpha - 1 \), we get:
\[ \alpha = 1.066 \alpha - 1.066 \]
10. Simplify and solve for \( \alpha \):
\[ 0.066 \alpha = 1.066 \] \[ \alpha = \frac{1.066}{0.066} \approx 16.15 \]
11. Given the expected range of 16 to 16, \( \alpha = 16 \) fits within the required constraints.
Thus, the value of \( \alpha \) is 16.
Approach Solution -2
The correct answer is 16
\((0.9)A_1\sqrt{\frac{K}{0.9}}=(0.9+0.124)A_2\sqrt{\frac{K}{0.9+0.124}}\)
\(\frac{A_1}{A_2}=\sqrt{\frac{0.9+0.124}{0.9}}\)
\(=\sqrt{\frac{1.024}{0.9}}\)
\(=\frac{α}{α−1}= α=16\)
\(\therefore\) the value of α will be 16
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Concepts Used:
Simple Harmonic Motion
Simple Harmonic Motion is one of the most simple forms of oscillatory motion that occurs frequently in nature. The quantity of force acting on a particle in SHM is exactly proportional to the displacement of the particle from the equilibrium location. It is given by F = -kx, where k is the force constant and the negative sign indicates that force resists growth in x.
This force is known as the restoring force, and it pulls the particle back to its equilibrium position as opposing displacement increases. N/m is the SI unit of Force.
Types of Simple Harmonic Motion
Linear Simple Harmonic Motion:
When a particle moves to and fro about a fixed point (called equilibrium position) along with a straight line then its motion is called linear Simple Harmonic Motion. For Example spring-mass system
Conditions:
The restoring force or acceleration acting on the particle should always be proportional to the displacement of the particle and directed towards the equilibrium position.
- – displacement of particle from equilibrium position.
- – Restoring force
- - acceleration
Angular Simple Harmonic Motion:
When a system oscillates angular long with respect to a fixed axis then its motion is called angular simple harmonic motion.
Conditions:
The restoring torque (or) Angular acceleration acting on the particle should always be proportional to the angular displacement of the particle and directed towards the equilibrium position.
Τ ∝ θ or α ∝ θ
Where,
- Τ – Torque
- α angular acceleration
- θ – angular displacement