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Chapter 10: Problem 1390

A small spherical steel ball is placed at a distance slightly away from the center of a concave mirror having radius of curvature \(250 \mathrm{~cm}\). If the ball is released, it will now move on the curved surface. What will be the periodic time of this motion? Ignore frictional force and take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$. (A) \((\pi / 4) \mathrm{s}\) (B) \(\pi \mathrm{s}\) (C) \((\pi / 2) \mathrm{s}\) (D) \(2 \pi \mathrm{s}\)

Short Answer

Expert verified
The short answer is: The periodic time of the motion is \(\pi \mathrm{s}\) (Answer choice B).

Step by step solution

01

Determine the length of the pendulum

The length of the pendulum is equivalent to the radius of curvature of the concave mirror, which is given as \(250 \mathrm{~cm}\). Since we need to use the value of \(\mathrm{g}\) in \(\mathrm{m/s}^2\), we will convert the length \(L\) into meters. So, \(L = 2.5 \mathrm{~m}\).
02

Use the formula for finding the periodic time of a simple pendulum

For a simple pendulum, the periodic time \(T\) can be given by the formula: \[T = 2\pi\sqrt{\frac{L}{g}}\] where \(L\) is the length (radius of curvature) in meters and \(g\) is the acceleration due to gravity in \(\mathrm{m/s}^2\).
03

Calculate the periodic time

Plug in the given values and calculate the periodic time: \[T = 2\pi\sqrt{\frac{2.5}{10}}\] \[T = 2\pi\sqrt{\frac{1}{4}}\] \[T = 2\pi\left(\frac{1}{2}\right)\] \[T = \pi \mathrm{s}\] The periodic time of the motion is \(\pi \mathrm{s}\), which corresponds to the answer choice (B).

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Most popular questions from this chapter

For particles \(\mathrm{A}\) and \(\mathrm{B}\) executing S.H.M., the equation for displacement is given by $\mathrm{y}_{1}=0.1 \sin (100 \mathrm{t}+\mathrm{p} / 3)$ and \(\mathrm{y}_{2}=0.1\) cos pt respectively. The phase difference between velocity of particle \(\mathrm{A}\) with respect to that of \(\mathrm{B}\) is \(\ldots \ldots\) \((\mathrm{A})-(\pi / 3)\) (B) \((\pi / 6)\) (C) \(-(\pi / 6)\) (D) \((\pi / 3)\)

Which of the following functions represents a travelling wave? (A) \((\mathrm{x}-\mathrm{vt})^{2}\) (B) in \((\mathrm{x}+\mathrm{vt})\) (C) \(\mathrm{e}^{-(\mathrm{x}+\mathrm{vt}) 2}\) (D) \(\\{1 /(\mathrm{x}+\mathrm{vt})\\}\)

The ratio of force constants of two springs is \(1: 5\). The equal mass suspended at the free ends of both springs are performing S.H.M. If the maximum acceleration for both springs are equal, the ratio of amplitudes for both springs is \(\ldots \ldots\) (A) \((1 / \sqrt{5})\) (B) \((1 / 5)\) (C) \((5 / 1)\) (D) \((\sqrt{5} / 1)\)

Two wires made up of same material are of equal lengths but their radii are in the ratio \(1: 2\). On stretching each of these two strings by the same tension, the ratio between their fundamental frequency is \(\ldots \ldots \ldots .\) (A) \(1: 2\) (B) \(2: 1\) (C) \(1: 4\) (D) \(4: 1\)

A person standing in a stationary lift measures the periodic time of a simple pendulum inside the lift to be equal to \(\mathrm{T}\). Now, if the lift moves along the vertically upward direction with an acceleration of $(\mathrm{g} / 3)$, then the periodic time of the lift will now be \((\mathrm{A}) \sqrt{3} \mathrm{~T}\) (B) \(\sqrt{(3 / 2) \mathrm{T}}\) (C) \((\mathrm{T} / 3)\) (D) \((\mathrm{T} / \sqrt{3})\)
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