To solve the problem of the suspended mass and the elongation of the springs, we need to analyze the behavior of springs under load. When a mass is attached to a spring, it stretches according to Hooke's Law, which states that the force exerted by a spring is proportional to its elongation. Let's break this down step by step.
Understanding Spring Mechanics
When a mass \( m \) is suspended from a spring, the force acting on the spring due to gravity is given by:
F = mg
where \( g \) is the acceleration due to gravity. According to Hooke's Law, the elongation \( x \) of the spring can be expressed as:
F = kx
Here, \( k \) is the spring constant, which measures the stiffness of the spring. Setting these two equations equal gives:
mg = kx
Initial Setup with One Spring
From the above equation, we can express the elongation of the spring when the mass is suspended:
x = \frac{mg}{k}
Cutting the Spring into Two Parts
Now, when the spring is cut into two parts, let’s denote the lengths of the larger and smaller springs as \( L_1 \) and \( L_2 \), respectively. The spring constants for these two parts will be different, as they depend on their lengths:
k_1 = \frac{k}{L_1} \quad \text{and} \quad k_2 = \frac{k}{L_2}
When the mass is suspended from the combination of these two springs in series, the effective spring constant \( k_{eff} \) can be calculated as:
\frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2}
Effective Spring Constant Calculation
Substituting the expressions for \( k_1 \) and \( k_2 \) gives:
\frac{1}{k_{eff}} = \frac{L_1}{k} + \frac{L_2}{k}
Thus, we can simplify this to:
k_{eff} = \frac{k}{\frac{L_1 + L_2}{k}} = \frac{k}{L_1 + L_2}
New Elongation with Combined Springs
When the mass is again suspended from the combination of springs, the elongation \( x' \) can be expressed as:
x' = \frac{mg}{k_{eff}} = \frac{mg(L_1 + L_2)}{k}
We know from the problem statement that this new elongation \( x' \) is equal to \( 4.5x \):
x' = 4.5x
Setting Up the Equation
Substituting for \( x \) gives:
\frac{mg(L_1 + L_2)}{k} = 4.5 \cdot \frac{mg}{k}
We can cancel \( mg/k \) from both sides, leading to:
L_1 + L_2 = 4.5
Finding the Ratio of Lengths
Let’s denote the ratio of the lengths of the larger spring to the smaller spring as:
r = \frac{L_1}{L_2}
From the equation \( L_1 + L_2 = 4.5 \), we can express \( L_1 \) in terms of \( L_2 \):
L_1 = rL_2
Substituting this into the length equation gives:
rL_2 + L_2 = 4.5
L_2(r + 1) = 4.5
Thus, we can express \( L_2 \) as:
L_2 = \frac{4.5}{r + 1}
Now substituting back for \( L_1 \):
L_1 = r \cdot \frac{4.5}{r + 1} = \frac{4.5r}{r + 1}
Final Ratio Calculation
Now we can find the ratio \( \frac{L_1}{L_2} \):
\frac{L_1}{L_2} = \frac{\frac{4.5r}{r + 1}}{\frac{4.5}{r + 1}} = r
Thus, the ratio of the lengths of the bigger spring to the smaller spring is simply \( r \), which can be derived from the elongation relationship. Given the elongation conditions, we can conclude that:
r = 4.5 - 1 = 3.5
Therefore, the ratio of the length of the bigger spring to the length of the smaller spring is:
3.5:1
