/*! 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 motion (SHM) h... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 1

A simple harmonic motion (SHM) has an amplitude \(A\) and time period \(T\). The time required by it to travel from \(x=A\) to \(x=A / 2\) is (A) \(T / 6\) (B) \(T / 4\) (C) \(T / 3\) (D) \(T / 2\)

Short Answer

Expert verified
The time required for the simple harmonic motion to travel from \(x=A\) to \(x=A/2\) is \(\frac{T}{6}\). The correct answer is (A).

Step by step solution

01

Identify the SHM displacement formula

The formula for the displacement of simple harmonic motion is given by: \[x(t) = A \cos(\omega t + \phi)\] where \(x(t)\) represents the displacement at time t, A is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase angle.
02

Find the equation relating time period and angular frequency

For SHM, the angular frequency \(\omega\) and the time period T are related by: \[\omega = \frac{2 \pi}{T} \]
03

Write the given conditions in terms of displacement formula

We have two conditions: 1. At \(t=0\), \(x(0) = A\). This implies that: \[A \cos(\phi) = A\] 2. At time \(t_1\), \(x(t_1) = A/2\). This implies that: \[A \cos(\frac{2 \pi}{T}t_1 + \phi) = A/2\]
04

Solve for the phase angle \(\phi\)

From the first condition, we can find the phase angle: \[\cos(\phi) = 1 \] \[\phi = 0\]
05

Substitute the values into the second condition and solve for \(t_1\)

Substituting the values of \(A\), \(\omega\), and \(\phi\) in the second condition, we get: \[A \cos(\frac{2 \pi}{T}t_1) = A/2\] \[\cos(\frac{2 \pi}{T}t_1) = \frac{1}{2}\] Now, we'll find the value of \(t_1\): \[\frac{2 \pi}{T}t_1 = \cos^{-1}\frac{1}{2}\] \[t_1 = \frac{T}{2 \pi}\cos^{-1}\frac{1}{2}\]
06

Calculate the value of \(t_1\)

\[\cos^{-1}\frac{1}{2} = \frac{\pi}{3}\] \[t_1 = \frac{T}{2 \pi}\times \frac{\pi}{3}\] \[t_1 = \frac{T}{6}\] So, the time required for the SHM to travel from x=A to x=A/2 is \(T / 6\). The correct answer is (A).

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Most popular questions from this chapter

Two SHMs are represented by the equations: \(Y_{1}=10 \sin [3 \pi t+\pi / 4]\) \(Y_{2}=5 \cos \pi t\) (A) The amplitude ratio of the two SHM is \(1: 1\). (B) The amplitude ratio of the two SHM is \(2: 1\). (C) Time periods of both the SHMs are equal. (D) Time periods of two SHMs are different.

A long wire \(A B C\) is made by joining two wires \(A B\) and \(B C\) of equal area of cross-section. \(A B\) has length \(4.8 \mathrm{~m}\) and mass \(0.12 \mathrm{~kg}\) while \(B C\) has length \(2.56\) \(\mathrm{m}\) and mass \(0.4 \mathrm{~kg}\). The wire is under a tension of \(160 \mathrm{~N}\). A wave \(Y\) (in \(\mathrm{cm})=3.5 \sin (k x-w t)\) is sent along \(A B C\) from end \(A\). No power is dissipated during propagation of wave. Column-I (A) Amplitude of reflected wave (B) Amplitude of transmitted wave (C) Maximum displacement of antinodes in the wire \(A B\) (D) Percentage fraction of power transmitted in the wire \(B C\) \begin{tabular}{l} Column-II \\ \hline 1. \(2.0\) \end{tabular} \(1.5\) 2 3 . 5 \(-1\) 82 4 \(=\) 5 . 92

The track followed for two perpendicular SHMs is a perfect ellipse when \(\left(\delta\right.\)-phase difference, \(A_{1}, A_{2}\) amplitudes) (A) \(\delta=\frac{\pi}{4}, A_{1} \neq A_{2}\) (B) \(\delta=\frac{3 \pi}{4}, A_{1} \neq A_{2}\) (C) \(\delta=\frac{\pi}{2}, A_{1} \neq A_{2}\) (D) \(\delta=\pi, A_{1}=A_{2}\)

The area of region covered by the TV broadcast by a TV tower of \(100 \mathrm{~m}\) height will be (radius of the earth \(=6.4 \times 10^{6} \mathrm{~m}\) ) (A) \(12.8 \pi \times 10^{8} \mathrm{~km}^{2}\) (B) \(1.28 \pi \times 10^{3} \mathrm{~km}^{2}\) (C) \(0.64 \pi \times 10^{3} \mathrm{~km}^{2}\) (D) \(1.28 \times 10^{3} \mathrm{~km}^{2}\)

A child swinging on a swing in sitting position, stands up, then the time period of the swing will (A) increase. (B) decrease. (C) remains same. (D) increase of the child is long and decreases if the child is short.
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