Chapter 15: Problem 12
The time period of a particle in simple harmonic motion is \(8 \mathrm{~s}\). At \(t=0\), it is at the mean position. The ratio of the distances travelled by it in the first and second, seconds is (a) \(1 / 2\) (b) \(1 / \sqrt{2}\) (c) \(1 /(\sqrt{2}-1)\) (d) \(1 / \sqrt{3}\)
Short Answer
Expert verified
The ratio is (c) \(1/(\sqrt{2} - 1)\).
Step by step solution
01
Understanding the problem
We need to find the ratio of the distances traveled by a particle in simple harmonic motion (SHM) in the first second and the second second. We know that the time period is 8 seconds and at time \(t=0\), it is at the mean position.
02
Formula for displacement in SHM
For a particle in SHM, the displacement \(x\) as a function of time \(t\) is given by \(x = A \sin\left(\frac{2\pi}{T}t\right)\), where \(A\) is the amplitude and \(T\) is the time period.
03
Calculate displacement at \(t=0\) and \(t=1s\)
When \(t=0\), \(x = A \sin(0) = 0\). When \(t=1\), \(x(1) = A \sin\left(\frac{2\pi}{8}\times 1\right) = A \sin\left(\frac{\pi}{4}\right) = \frac{A}{\sqrt{2}}\).
04
Calculate displacement at \(t=2s\)
When \(t=2\), \(x(2) = A \sin\left(\frac{2\pi}{8}\times 2\right) = A \sin\left(\frac{\pi}{2}\right) = A\).
05
Calculate distances traveled in 1st and 2nd seconds
The distance traveled in the first second is from \(t=0\) to \(t=1\), which is \(|x(1) - x(0)| = \frac{A}{\sqrt{2}} - 0 = \frac{A}{\sqrt{2}}\). The distance traveled in the second second is from \(t=1\) to \(t=2\), which is \(|x(2) - x(1)| = A - \frac{A}{\sqrt{2}}\).
06
Find the ratio of distances
Now, calculate the ratio of the distance traveled in the first second to that in the second second: \(\text{Ratio} = \frac{\frac{A}{\sqrt{2}}}{A - \frac{A}{\sqrt{2}}} = \frac{1/\sqrt{2}}{1 - 1/\sqrt{2}} = \frac{1}{\sqrt{2} - 1}.\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Time Period
The time period in simple harmonic motion (SHM) is a crucial concept. It represents the duration required for a particle to complete one full cycle of motion. In SHM, a particle moves back and forth in a regular, repeating path. This period of oscillation is constant for a given system.
Understanding the time period helps us predict how a particle will move over time. For example, if a particle has an 8-second time period, it takes 8 seconds to move from its starting point, through all intermediary points, back to its starting point again.
Key points to remember about the time period:
- The time period (T) is typically measured in seconds.
- In SHM, each cycle of motion through the equilibrium point and back completes in this time frame.
- It's influenced by the system's physical properties, like mass and spring constant in a mass-spring system or the length of a pendulum.
Displacement in SHM
Displacement in simple harmonic motion refers to the position of a particle relative to its equilibrium or mean position at any point in time. It is crucial to understanding how the particle travels over time. By using a mathematical equation, we can determine this displacement.For a particle in SHM, the displacement, denoted by \(x(t)\), at a time \(t\) is given by:\[x(t) = A \sin\left(\frac{2\pi}{T}t\right)\ \]where:
- \(A\) is the amplitude, representing the maximum displacement from the mean position.
- \(T\) is the time period, indicating how long it takes to complete one cycle of motion.
- \(t\) is the time at which the displacement is measured.
Amplitude
Amplitude in simple harmonic motion is the maximum extent of displacement from the mean position. It is a fundamental characteristic of SHM that indicates how far the particle can move away from its equilibrium point during oscillations.Several essential features of amplitude include:
- It represents the peak (or dip) value of the displacement during one complete cycle of motion.
- Amplitude is always positive and is typically denoted as \(A\).
- In practical problems, it sets a boundary for the particle's oscillation, acting as a constant in the displacement formula.