On a frictionless horizontal plane, a bob of mass m = 0.1 kg is attached to a spring with natural le

Q: Why do we take half the time period for each side?

Because the block only completes half a cycle (from extreme to mean to extreme) on one side before crossing over to the other side where the spring constant changes.

Q: Does the amplitude on the left side equal the amplitude on the right side?

No! Since energy is conserved, $\frac{1}{2}k_1 A_1^2 = \frac{1}{2}k_2 A_2^2$. The stiffer side will have a smaller amplitude.

Q: Why doesn't the different amplitude affect the time period?

In Simple Harmonic Motion, the time period is fundamentally independent of the amplitude. It only depends on the mass and the spring constant.

Q: What if the block was released from $l = 0.05$ m instead?

The block would still oscillate, but it would start in the compressed region. The time period would remain exactly the same as long as it has enough energy to cross the mean position.

Q: Is the motion still considered Simple Harmonic Motion?

Strictly speaking, the entire motion is not a single SHM because the restoring force is not linearly proportional to displacement across the entire range. However, it is composed of two perfect halves of different SHMs.

Question

On a frictionless horizontal plane, a bob of mass $m = 0.1$ kg is attached to a spring with natural length $l_0 = 0.1$ m. The spring constant is $k_1 = 0.009 	ext{ Nm}^{-1}$ when the length of the spring $l > l_0$ and is $k_2 = 0.016 	ext{ Nm}^{-1}$ when $l < l_0$. Initially the bob is released from $l = 0.15$ m. Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is $T = (n\pi)$ s, then the integer closest to n is ________.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are observing a classic spring-mass system on a perfectly frictionless horizontal plane. At first glance, it looks like a textbook Simple Harmonic Motion (SHM) problem. But there is a fascinating twist! The spring possesses a "split personality."

When the spring is stretched beyond its natural length ($l > l_0$), it behaves with a stiffness of $k_1 = 0.009 	ext{ N/m}$. However, the moment it compresses ($l < l_0$), it suddenly becomes stiffer, operating with a new spring constant $k_2 = 0.016 	ext{ N/m}$.

Because the restoring force changes depending on which side of the mean position the block is on, the motion is not a single, uniform SHM. Instead, it is a combination of two different half-SHMs seamlessly stitched together at the mean position.

### The Master Equation

To find the total time period $T$ of one full oscillation, we must break the journey into two distinct halves.

A standard spring-mass system has a time period given by $T = 2\pi\sqrt{\frac{m}{k}}$. In our piecewise system, the block spends exactly half of a theoretical full cycle in the stretched region, and the other half in the compressed region.

Therefore, the time spent in the stretched region is:
$$T_{	ext{stretch}} = \frac{1}{2} \left( 2\pi\sqrt{\frac{m}{k_1}} ight) = \pi\sqrt{\frac{m}{k_1}}$$

Similarly, the time spent in the compressed region is:
$$T_{	ext{compress}} = \frac{1}{2} \left( 2\pi\sqrt{\frac{m}{k_2}} ight) = \pi\sqrt{\frac{m}{k_2}}$$

The total time period is simply the sum of these two durations:
$$T = \pi\sqrt{\frac{m}{k_1}} + \pi\sqrt{\frac{m}{k_2}}$$

### Final Calculation

Now, let's substitute the given values into our master equation. We know the mass $m = 0.1 	ext{ kg}$, $k_1 = 0.009 	ext{ N/m}$, and $k_2 = 0.016 	ext{ N/m}$.

$$T = \pi\sqrt{\frac{0.1}{0.009}} + \pi\sqrt{\frac{0.1}{0.016}}$$

To avoid silly mistakes, let's eliminate the decimals by multiplying the numerator and denominator inside the square roots by $1000$:

$$T = \pi\sqrt{\frac{100}{9}} + \pi\sqrt{\frac{100}{16}}$$

This simplifies beautifully! Taking the square roots gives:

$$T = \pi\left(\frac{10}{3} + \frac{10}{4}ight)$$

Taking the LCM of $3$ and $4$, which is $12$:

$$T = \pi\left(\frac{40 + 30}{12}ight) = \frac{70}{12}\pi$$

Dividing $70$ by $12$ yields approximately $5.83$.

$$T \approx 5.83\pi$$

The question states that the time period is $T = (n\pi)$ seconds and asks for the integer closest to $n$. Comparing our result, we find $n \approx 5.83$. The closest integer to $5.83$ is **$6$**.

### The Way Forward

This is a favorite concept for JEE because it tests your fundamental understanding of time periods. Notice how the initial release position ($l = 0.15 	ext{ m}$) was completely irrelevant to our calculation!

In SHM, the time period is strictly independent of the amplitude. Even though the block will compress the spring by a different amount than it stretched it (due to energy conservation $\frac{1}{2}k_1 A_1^2 = \frac{1}{2}k_2 A_2^2$), the time it takes to complete the journey remains locked to the mass and the spring constants.

Enter Numerical Value:

Visualized Solution

$m = 0.1 kg, l_{0} = 0.1 m$

$T = T_{stretch} + T_{compress}$

$T = π \frac{m}{k _{1}} + π \frac{m}{k _{2}}$

$T = π \frac{0.1}{0.009} + π \frac{0.1}{0.016}$

$T = π \frac{100}{9} + π \frac{100}{16}$

$T = π (\frac{10}{3} + \frac{10}{4})$

$T = \frac{70}{12} π$

$n \approx 6$

$Amplitude Independence$

The Sigma Insight: Simple Harmonic Motion (SHM)

Analyzing the Setup

The Master Equation

Final Calculation

The Way Forward

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m=0.1 kg,l0​=0.1 m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=Tstretch​+Tcompress​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=πk1​m​​+πk2​m​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=π0.0090.1​​+π0.0160.1​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=π9100​​+π16100​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=π(310​+410​)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=1270​π

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n≈6

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