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Assertion The percentage change in time period is \(1.5 \%\), if the length of simple pendulum increases by \(3 \%\). Reason Time period is directly proportional to length of pendulum.

Short Answer

Expert verified
The assertion is correct; the percentage increase in time period is approximately 1.5%.

Step by step solution

01

Identify the Relationship

The time period of a simple pendulum is given by the formula \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( T \) is the time period, \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity.
02

Establish Proportional Relationship

Since the equation involves a square root function, the time period \( T \) is proportional to the square root of the length \( L \), i.e., \( T \propto \sqrt{L} \). This indicates that if the length \( L \) changes by a certain percentage, the change in the time period will be the square root of that percentage change.
03

Calculate Percentage Change in Length

The problem states that the length of the pendulum increases by 3%. This means an increase of 0.03 in decimal form.
04

Apply the Change to the Time Period

To find the percentage change in time period, we use \( T_{\text{new}} = T_{\text{original}} \times \sqrt{1.03} \). \(sqrt{1.03} \approx 1.014889\), so the increase in the time period is approximately 1.4889%.
05

Conclusion and Comparison

The calculated percentage increase in the time period, 1.4889%, is close to the given assertion of 1.5%. Hence, the assertion is approximately correct, and the reason provided is true as the time period is indeed directly proportional to the square root of the pendulum length.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Time Period of a Simple Pendulum
When you think about a pendulum, you've probably seen one gently swinging back and forth in a grandfather clock or a science demonstration. The time it takes for the pendulum to complete a full swing and return to its starting point is called the "time period." Understanding the time period of a pendulum is crucial in grasping the underlying physics of its motion.

The time period of a simple pendulum is dependent on two main factors: the length of the pendulum and the acceleration due to gravity. It is mathematically represented as \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( T \) is the time period, \( L \) represents the length of the pendulum, and \( g \) is the acceleration due to gravity.

This formula reveals that the time period is directly related to the length of the pendulum. If you increase the length, the pendulum swings more slowly, thus increasing its time period. Conversely, a shorter pendulum swings more quickly. Gravity, interestingly, stays constant for a location on Earth, so it's the length that will mainly affect the period.
Exploring Proportional Relationships in Pendulum Motion
Proportional relationships are a cornerstone of understanding pendulum motion. In the context of a simple pendulum, we see a special type of proportionality.

The time period \( T \) is proportional to the square root of the length \( L \) of the pendulum. Mathematically, it is expressed as \( T \propto \sqrt{L} \). This means if you know by how much the length changes, you can determine how much the time period will change by taking the square root of the change in length.

This relationship can be understood like this:
  • Increase the length of the pendulum, and the time period will increase, but not one-to-one. Instead, it follows the pattern of a square root. For instance, if the length increases by 9%, the time period increases by \( \sqrt{.09} \), which is 3% (not 9%).
  • This proportional relationship allows predictions about how changes in length affect the time period, which is useful for engineering and scientific experiments.
Understanding this helps realize that the changes are not linear but rather exhibit a diminishing or amplifying effect depending on whether the length is being increased or decreased.
Calculating and Understanding Percentage Change in Time Period
Understanding percentage change can assist in predicting how small alterations affect the time period of a pendulum. Firstly, percentage change is a way of expressing how much one quantity has increased or decreased relative to its original value.

In the case of a simple pendulum, if the length changes, there's a need to calculate how this will affect the time period. As mentioned before, the time period \( T \) is proportional to the square root of the length \( L \).

For example, if the length of a pendulum increases by 3%, you express this as an increase expressed by \( (1 + 0.03) \). To find the effect on the time period, you compute \( T_{\text{new}} = T_{\text{original}} \times \sqrt{1.03} \). Calculating \( \sqrt{1.03} \approx 1.014889 \) shows a 1.4889% increase in the time period.

This slight change demonstrates the non-linear nature of how the time period adapts to changes in the pendulum’s length. Recognizing and accurately calculating these percentage changes are essential steps in working with pendulums, ensuring precise timing in clocks and experiments.

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Most popular questions from this chapter

A particle, with restoring force proportional to displacement and resisting force proportional to velocity is subjected to a force, \(F=F_{0} \sin \omega t\) If, the amplitude of the particle is maximum for \(\omega=\omega_{1}\) and the energy of the particle is maximum for \(\omega=\omega_{2}\), then (a) \(\omega_{1}=\omega_{0}\) and \(\omega_{2} \neq \omega_{0}\) (b) \(\omega_{1}=\omega_{0}\) and \(\omega_{2}=\omega_{0}\) (c) \(\omega_{1} \neq \omega_{0}\) and \(\omega_{2}=\omega_{0}\) (d) \(\omega_{1} \neq \omega_{0}\) and \(\omega_{2} \neq \omega_{0}\)

Two particles are executing simple harmonic motion of the same amplitude \(A\) and frequency \(\omega\) along the \(x\)-axis. Their mean position is separated by distance \(X_{0}\left(X_{0}>A\right)\). If the maximum separation between them is \(\left(X_{0}+A\right)\) the phase difference during their motion is (a) \(\frac{\pi}{2}\) (b) \(\frac{\pi}{3}\) (c) \(\frac{\pi}{4}\) (d) \(\frac{\pi}{6}\)

The acceleration due to gravity on the surface of moon is \(1.7 \mathrm{~m} / \mathrm{s}^{2}\). What is the time period of a simple pendulum on the surface of moon, if its time period on the surface of earth is \(3.5 \mathrm{~s}\) ? \(\left(g\right.\) on the surface of earth is \(9.8 \mathrm{~m} / \mathrm{s}^{2} .\) ) \(\quad\) [NCERT] (a) \(6.4 \mathrm{~s}\) (b) \(7.4 \mathrm{~s}\) (c) \(9.4 \mathrm{~s}\) (d) \(8.4 \mathrm{~s}\)

The total energy of a simple harmonic oscillator is proportional to [Kerala CET 2008] (a) square root of displacement (b) velocity (c) frequency (d) amplitude (e) square of the amplitude

A uniform cylinder of length \(L\) and mass \(M\) having cross-sectional area \(A\) is suspended with its vertical length, from a fixed point by a massless spring, such that it is half submerged in a liquid of density \(d\) at equilibrium position. When released, it starts oscillating vertically with a small amplitude. If the force constant of the spring is \(k\), the frequency of oscillation of the cylinder is $$ \frac{1}{2 \pi}\left(\frac{k-A d g}{M}\right)^{1 / 2} \quad \text { (b) } \frac{1}{2 \pi}\left(\frac{k+A d g}{M}\right)^{1 / 2} $$ (c) \(\frac{1}{2 \pi}\left(\frac{k-d g L}{M}\right)^{1 / 2}\) (d) \(\frac{1}{2 \pi}\left(\frac{k+A g L}{A d g}\right)^{1 / 2}\)

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