Mass suspended to a spring is pulled down by 2.5 cm and let go. The mass oscillates with an amplitude of

Options

A. 2.5 cm
B. 5 cm
C. 7.5 cm
D. 10 cm

Answer

To solve the problem, we need to determine the amplitude of the oscillation when a mass is pulled down by 2.5 cm and then released. Here's a step-by-step solution:

### Step 1: Understand the System
We have a mass suspended from a spring. When the mass is at its natural length, it is in equilibrium due to the balance between the gravitational force acting downwards (weight of the mass) and the spring force acting upwards.

**Hint:** Identify the forces acting on the mass when it is at rest.

### Step 2: Define the Equilibrium Position
When the mass is hanging freely, it stretches the spring to a new equilibrium position (let's call this position \( x_0 \)). At this position, the spring force \( kx_0 \) (where \( k \) is the spring constant) balances the weight of the mass \( mg \).

**Hint:** Use the equation \( kx_0 = mg \) to relate the spring constant, mass, and the equilibrium position.

### Step 3: Pulling the Mass Down
The problem states that the mass is pulled down by 2.5 cm from the equilibrium position and then released. This means that the new position of the mass is 2.5 cm below the equilibrium position.

**Hint:** Recognize that pulling the mass down creates a displacement from the equilibrium position.

### Step 4: Define the Amplitude
In oscillatory motion, the amplitude is defined as the maximum displacement from the mean (equilibrium) position. Since the mass is pulled down by 2.5 cm and then released, this distance represents the maximum displacement from the equilibrium position.

**Hint:** Remember that the amplitude is the maximum distance from the mean position in oscillations.

### Step 5: Conclusion
Thus, the amplitude of the oscillation is equal to the distance the mass was pulled down from the equilibrium position, which is 2.5 cm.

**Final Answer:** The mass oscillates with an amplitude of **2.5 cm**.