Question

Two springs with constants k and 2k are connected in series and a mass m is hanged. The time period of oscillation is

Hint:

For springs in series, the equivalent spring constant is always less than the smallest individual spring constant. $k_{eq} = \frac{\text{product}}{\text{sum}} = \frac{k \cdot 2k}{k + 2k} = \frac{2k^2}{3k} = \frac{2k}{3}$.

Answer 1

Step 1: Understanding the Concept:
When springs are connected in series, the same force (tension) acts on each spring, but their extensions add up. This arrangement can be replaced by a single equivalent spring with an effective spring constant. The time period of a mass-spring system depends on mass and this effective spring constant.

Step 2: Key Formula or Approach:
1. The equivalent spring constant ($k_{eq}$) for two springs ($k_1, k_2$) in series is:
\[ \frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} \]
2. The time period ($T$) of the oscillating mass is:
\[ T = 2\pi\sqrt{\frac{m}{k_{eq}}} \]

Step 3: Detailed Explanation:
Given spring constants: $k_1 = k$ and $k_2 = 2k$.
Calculate the equivalent spring constant $k_{eq}$:
\[ \frac{1}{k_{eq}} = \frac{1}{k} + \frac{1}{2k} \]
Find a common denominator:
\[ \frac{1}{k_{eq}} = \frac{2}{2k} + \frac{1}{2k} = \frac{3}{2k} \]
Inverting both sides gives:
\[ k_{eq} = \frac{2k}{3} \]
Now, substitute $k_{eq}$ into the time period formula:
\[ T = 2\pi\sqrt{\frac{m}{k_{eq}}} \]
\[ T = 2\pi\sqrt{\frac{m}{\frac{2k}{3}}} \]
\[ T = 2\pi\sqrt{\frac{3m}{2k}} \]

Step 4: Final Answer:
The time period of oscillation is $2\pi\sqrt{\frac{3m}{2k}}$.