/*! 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 1 A simple harmonic oscillator has... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 1

A simple harmonic oscillator has an amplitude \(A\) and time period \(T\). The time required by it to travel from \(x=A\) to \(x=A / 2\) is \(\quad\) [CBSE 1992; SCRA 1996] (a) \(T / 6\) (b) \(T / 4\) (c) \(T / 3\) (d) \(T / 2\)

Short Answer

Expert verified
The time required by the oscillator to travel from \(x=A\) to \(x=A / 2\) is \(T / 6\). Hence, the correct answer is option (a) \(T / 6\).

Step by step solution

01

Write down the equation of motion for SHM

The equation of motion for a simple harmonic oscillator can be represented as \(x = A \cos (蠅t + 蠁)\) where \(x\) is the displacement, \(A\) the amplitude, \(蠅\) the angular frequency, \(t\) time and \(蠁\), the phase constant. If the oscillator starts from amplitude \(A\) and \(蠁 = 0\), the equation becomes \(x = A \cos 蠅t\).
02

Use the equation to find time

Since we are looking for the time requires to go from \(x = A\) to \(x = A / 2\), we can use the equation of motion. At \(x=A\), \(t=0\). At \(x = A/2\), we can solve the equation \(A / 2 = A \cos 蠅t\). Dividing both sides by \(A\), we get \(\cos 蠅t = 1 / 2\). The value of 蠅t for which cos 蠅t = 1 / 2 for the first time is \(蟺 / 3\).
03

Find the Time Period

The time period \(T\) is related to the angular frequency \(蠅\) by the equation \(蠅 = 2蟺 / T\). Substituting this into \(蠅t = 蟺 / 3\), we get \(t = T / 6\) after solving the equation.

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Most popular questions from this chapter

A mass \(m\) is suspended from a string of length \(l\) and force constant \(k\). The frequency of vibration of the mass is \(f_{1}\). The spring is cut in to two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of mass is \(f_{2}\). Which of the following reaction between the frequencies is correct. [NCERT \(198_{3}\); CPMT 1986; MP PMT 1991] (a) \(f_{1}=\sqrt{2} f_{2}\) (b) \(f_{1}=f_{2}\) (c) \(f_{1}=2 f_{2}\) (d) \(f_{2}=\sqrt{2} f_{1}\)

A particle of mass 10 grams is executing S.H.M. with an amplitude of o.5 meter and circular frequenc of 10 radian \(/ \mathrm{sec}\). The maximum value of the force acting on the particle during the course of oscillatio: is [MP PMT 200 (a) \(25 N\) (b) \(5 N\) (c) \(2.5 N\) (d) \(0.5 N\)

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