If two similar springs each of spring constant K1 are joined in series, the new spring constant and

Q: Why do springs in series have a smaller equivalent spring constant?

When springs are connected in series, the same force is applied to both, causing each to stretch individually. The total extension is the sum of their individual extensions. More extension for the same force means the system is effectively "stretchier" or less stiff, resulting in a lower spring constant.

Q: What would happen if the springs were connected in parallel instead?

In a parallel combination, the springs share the applied force, and they stretch by the same amount. The equivalent spring constant is the sum of the individual spring constants ($k_{eq} = k_1 + k_2$), making the system stiffer.

Q: Does the mass $m$ affect the factor by which the time period changes?

No, the mass $m$ remains constant. When you take the ratio of the new time period to the old time period, the mass $m$ cancels out completely. The change factor depends solely on the ratio of the spring constants.

Q: Why is the formula for springs in series similar to capacitors in series but opposite to resistors?

The spring constant $k$ relates force to extension ($F=kx$). In series, extensions add up ($x = x_1 + x_2$) for the same force, leading to $F/k_{eq} = F/k_1 + F/k_2$. This mathematical structure is identical to capacitors in series ($V = V_1 + V_2 \Rightarrow Q/C_{eq} = Q/C_1 + Q/C_2$), where voltage drops add up for the same charge.

Q: Can I just add the time periods of the individual springs to get the new time period?

No, the time period is proportional to the inverse square root of the spring constant ($1/\sqrt{k}$). You cannot simply add the time periods. You must first find the equivalent spring constant and then calculate the new time period using the formula.

Question

If two similar springs each of spring constant $K_1$ are joined in series, the new spring constant and time period would be changed by a factor

Options:

(A) $\frac{1}{2}, \sqrt{2}$

(B) $\frac{1}{4}, \sqrt{2}$

(C) $\frac{1}{4}, 2\sqrt{2}$

(D) $\frac{1}{2}, 2\sqrt{2}$

Sigma Percentile · Accepted Answer

The problem asks us to determine how the equivalent spring constant and the time period of oscillation change when two identical springs are connected in series.

### The Series Connection
Imagine you have a spring hanging from a rigid support. Now, take another identical spring and attach it directly to the bottom of the first one. This end-to-end arrangement is known as a **series combination**.

When springs are connected in series, any force applied to the bottom of the system is transmitted equally through both springs. Because both springs stretch under this force, the total extension of the system is the sum of their individual extensions. This makes the overall system "stretchier" or less stiff than a single spring.

### Calculating the Equivalent Spring Constant
To quantify this new stiffness, we use the formula for the equivalent spring constant ($k_{eq}$) of springs in series. The relationship is reciprocal:
$$ \frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} $$

In our specific case, both springs are identical, meaning $k_1 = k_2 = K_1$. Substituting this into our formula gives:
$$ \frac{1}{k_{eq}} = \frac{1}{K_1} + \frac{1}{K_1} $$

$$ \frac{1}{k_{eq}} = \frac{2}{K_1} $$

To find $k_{eq}$, we simply take the reciprocal of both sides:
$$ k_{eq} = \frac{K_1}{2} $$

This tells us that the new spring constant is exactly half of the original. Therefore, the spring constant changes by a factor of **$\frac{1}{2}$**.

### Impact on the Time Period
Now, let's see how this change affects the time period of oscillation if a mass $m$ is attached to the system. The standard formula for the time period $T$ of a spring-mass system is:
$$ T = 2\pi\sqrt{\frac{m}{k}} $$

For our new series system, the time period $T'$ will use the equivalent spring constant $k_{eq}$:
$$ T' = 2\pi\sqrt{\frac{m}{k_{eq}}} $$

Substitute $k_{eq} = \frac{K_1}{2}$ into the equation:
$$ T' = 2\pi\sqrt{\frac{m}{\frac{K_1}{2}}} $$

Because the $2$ is in the denominator of the fraction inside the square root, it flips up to the numerator:
$$ T' = 2\pi\sqrt{\frac{2m}{K_1}} $$

We can pull the $\sqrt{2}$ out to the front to clearly see the relationship with the original time period:
$$ T' = \sqrt{2} \left( 2\pi\sqrt{\frac{m}{K_1}} \right) $$

$$ T' = \sqrt{2} T $$

The new time period is $\sqrt{2}$ times the original time period. Thus, the time period changes by a factor of **$\sqrt{2}$**.

Combining both results, the factors are $\frac{1}{2}$ and $\sqrt{2}$, which corresponds to option (a).

Animated Solution for Physics - Oscillations: If two similar springs each of spring constant $K_{1}$ are joined in series, the new spring constant and time period would be changed by a factor

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Visualized Solution

$Springs in Series$

$Equivalent Spring Constant Formula$

$Calculating k_{e q}$

$Time Period Formula$

$Calculating New Time Period T^{'}$

$Final Conclusion$

The Sigma Insight: Simple Harmonic Motion (SHM)

The Series Connection Imagine you have a spring hanging from a rigid support

Calculating the Equivalent Spring Constant To quantify this new stiffness, we use the formula for the equivalent spring constant ( $k_{e q}$ ) of springs in series

Impact on the Time Period Now, let's see how this change affects the time period of oscillation if a mass $m$ is attached to the system

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Springs in Series

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equivalent Spring Constant Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating keq​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time Period Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating New Time Period T′

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Conclusion