/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 A simple pendulum has time perio... [FREE SOLUTION] | 91Ó°ÊÓ

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A simple pendulum has time period \(T_{1}\). The point of suspension is now moved upward according to the relation \(y=k t^{2}\left(k=1 \mathrm{~m} / \mathrm{s}^{2}\right)\), where \(y\) is the vertical displacement. The time period now becomes \(T_{2}\). The ratio of \(T_{1}^{-2} / T_{2}^{2}\) is (Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (a) \(6 / 5\) (b) \(5 / 6\) (c) 1 (d) \(4 / 5\)

Short Answer

Expert verified
The ratio \(\frac{T_{1}^{-2}}{T_{2}^{2}}\) is \(\frac{6}{5}\), so option (a) is correct.

Step by step solution

01

Identify the formula for the time period of a simple pendulum

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

Determine the length change due to vertical displacement

The suspension point is moved up according to \(y = k t^2\), so the effective length \(L'\) becomes \(L - y = L - k t^2\). Since \(k = 1 \, \text{m/s}^2\), \(y = t^2\).
03

Calculate the new time period with the adjusted length

The new time period \(T_2\) for length \(L - t^2\) is \(T_2 = 2\pi \sqrt{\frac{L - t^2}{g}}\).
04

Express the original time period

The original time period \(T_1\) is \(T_1 = 2\pi \sqrt{\frac{L}{g}}\).
05

Setting up the ratio of the squares of the inverse time periods

The question asks for \(\frac{T_1^{-2}}{T_2^2}\). \[ T_1^{-2} = \left(\frac{1}{2\pi}\right)^2 \left(\frac{g}{L}\right) \]\[ T_2^2 = \left(2\pi\right)^2 \left(\frac{L - t^2}{g}\right) \]
06

Simplify the expression for the ratio

Substitute the expressions for \(T_1^{-2}\) and \(T_2^2\) into the ratio:\[\frac{T_1^{-2}}{T_2^2} = \frac{\left(\frac{1}{2\pi}\right)^2 \left(\frac{g}{L}\right)}{\left(2\pi\right)^2 \left(\frac{L - t^2}{g}\right)} = \frac{g^2/L}{4\pi^4 (L - t^2)/g}\]This simplifies to:\[\frac{T_1^{-2}}{T_2^2} = \frac{g^2}{(L)(L-t^2)}\]As \(g = 10 \text{ m/s}^2\), substitute to check the change based on time.
07

Evaluate specific conditions

Assuming this is checked over a duration where \(t\) becomes such that \(L - t^2 = \frac{5}{6}L\), then:\[\frac{T_1^{-2}}{T_2^2} = \frac{g^2}{L \cdot \frac{5}{6}L} = \frac{6g^2}{5L^2} = \frac{6}{5}\]Thus, option (a) \(\frac{6}{5}\) is correct.

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Key Concepts

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

Time Period Calculation
When dealing with pendulums, understanding the time period calculation is crucial. The time period of a simple pendulum is the time it takes to complete one full swing back and forth.
The classic formula used in physics to determine this is given by:
  • \[ T = 2\pi \sqrt{\frac{L}{g}} \]
Here, \(T\) represents the time period, \(L\) is the pendulum's length, and \(g\) denotes the acceleration due to gravity, commonly approximated by \(10 \text{ m/s}^2\) on Earth's surface.This formula shows that the time period is primarily dependent on the length of the pendulum and gravity. Longer strings result in longer periods, while stronger gravitational fields decrease the period. It's important to note that the time period is independent of the mass of the pendulum bob, which is an interesting feature of harmonic motion.
Vertical Displacement Effect
Vertical displacement in a pendulum affects its effective length and thus its time period. In this exercise, the suspension point of the pendulum is vertically displaced according to the equation \(y = k t^2\), where \(k\) is a constant. As in our case, \(k = 1 \text{ m/s}^2\), simplifying the equation to \(y = t^2\).
The adjustment in length due to vertical displacement is given by:
  • The new effective length is \(L' = L - y\).
  • Substituting \(y = t^2\), we get \(L' = L - t^2\).
This new effective length impacts the pendulum's time period. As the length decreases due to the upward movement, the pendulum swings faster, altering the overall dynamics. Understanding this relationship helps in predicting how changes in the physical setup impact the pendulum's behavior.
Pendulum Length Adjustment
Adjusting the pendulum's length is directly related to its time period. In our problem, the pendulum's length effectively changes because of the vertical motion. This makes the pendulum swing differently.Considering the modified length \(L' = L - t^2\), we can calculate the new time period:
  • \[ T_2 = 2\pi \sqrt{\frac{L - t^2}{g}} \]
This formula allows us to understand how fluctuations in the physical set-up—like moving the suspension point—impact the period. Changing the length not only gives insight into How the pendulum will react but also illustrates the significant role of physical modifications in real-world applications. Fits perfectly if you're adjusting pendulum setups to maintain desired time periods for precise tasks.
Physics Problem Solving
Physics problem solving can be methodical and follows structured steps. Identifying knowns and unknowns is key. In this exercise, we began by identifying the known equation for the time period and altered it according to the physical changes presented by the displacement.
Steps to solve the problem were:
  • Establish the original formula for the time period.
  • Adjust it for changes in pendulum length due to vertical displacement.
  • Set up the mathematical relationship given in the problem: the ratio of the squares of inverse time periods \(\frac{T_1^{-2}}{T_2^2}\).
  • Simplify this expression using known values and assumptions.
These practices are beneficial in any scenario, given the structured approach can be applied to myriad physics problems—enhancing problem-solving efficiency and understanding of the underlying principles. By applying logical steps in order, complex physics problems become manageable, unveiling solutions piece by piece.

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