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Chapter 10: Problem 1373

The maximum velocity and maximum acceleration of a particle executing S.H.M. are \(1 \mathrm{~m} / \mathrm{s}\) and \(3.14 \mathrm{~m} / \mathrm{s}^{2}\) respectively. The frequency of oscillation for this particle is...... (A) \(0.5 \mathrm{~s}^{-1}\) (B) \(3.14 \mathrm{~s}^{-1}\) (C) \(0.25 \mathrm{~s}^{-1}\) (D) \(2 \mathrm{~s}^{-1}\)

Short Answer

Expert verified
The frequency of oscillation for this particle is \(0.5 \mathrm{s}^{-1}\), which corresponds to option (A).

Step by step solution

01

Write down the given information

The maximum velocity (v_max) is given as \(1 \mathrm{~m/s}\) and the maximum acceleration (a_max) is given as \(3.14 \mathrm{~m/s^2}\).
02

Write down the formula for maximum velocity and maximum acceleration in S.H.M.

In a simple harmonic motion, the maximum velocity (v_max) is given by the formula: \(v_{max} = \omega A\) Where \(\omega\) is the angular frequency and \(A\) is the amplitude of oscillation. Similarly, the maximum acceleration (a_max) is given by the formula: \(a_{max} = \omega^2 A\)
03

Find the relationship between maximum velocity and maximum acceleration

We can find the relationship between maximum velocity and maximum acceleration by dividing the two equations from Step 2. So we have: \(\frac{v_{max}}{a_{max}} = \frac{\omega A}{\omega^2 A}\) Solving for \(\omega\) we get: \(\omega =\frac{v_{max}}{a_{max}}\)
04

Calculate the angular frequency

Now that we have a relationship between maximum velocity and maximum acceleration, we can find the angular frequency \(\omega\) using the given values. So, \(\omega = \frac{1 \mathrm{~m/s}}{3.14 \mathrm{~m/s^2}} = \frac{1}{3.14}\mathrm{s^{-1}}\)
05

Calculate the frequency

Now that we have found the angular frequency, we can find the oscillation frequency (f) using the formula: \(f = \frac{\omega}{2\pi}\) Substituting the value of \(\omega\) into the formula we get: \(f = \frac{\frac{1}{3.14}\mathrm{s^{-1}}}{2\pi} = \frac{1}{2\pi \cdot 3.14} \mathrm{s}^{-1} = 0.5 \mathrm{s}^{-1}\) So, the frequency of oscillation for this particle is \(0.5 \mathrm{s}^{-1}\), which corresponds to option (A).

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Most popular questions from this chapter

As shown in figure, two light springs having force constants \(\mathrm{k}_{1}=1.8 \mathrm{~N} \mathrm{~m}^{-1}\) and $\mathrm{k}_{2}=3.2 \mathrm{~N} \mathrm{~m}^{-1}$ and a block having mass \(\mathrm{m}=200 \mathrm{~g}\) are placed on a frictionless horizontal surface. One end of both springs are attached to rigid supports. The distance between the free ends of the spring is \(60 \mathrm{~cm}\) and the block is moving in this gap with a speed \(\mathrm{v}=120 \mathrm{~cm} \mathrm{~s}^{-1}\).When the block is moving towards \(k_{1}\), what will be the time taken for it to get maximum compressed from point \(\mathrm{C}\) ? (A) \(\pi \mathrm{s}\) (B) \((2 / 3) \mathrm{s}\) (C) \((\pi / 3) \mathrm{s}\) (D) \((\pi / 4) \mathrm{s}\)

A body having mass \(5 \mathrm{~g}\) is executing S.H.M. with an amplitude of \(0.3 \mathrm{~m}\). If the periodic time of the system is $(\pi / 10) \mathrm{s}\(, then the maximum force acting on body is \)\ldots \ldots \ldots \ldots$ (A) \(0.6 \mathrm{~N}\) (B) \(0.3 \mathrm{~N}\) (C) \(6 \mathrm{~N}\) (D) \(3 \mathrm{~N}\)

Twenty four tuning forks are arranged in such a way that each fork produces 6 beats/s with the preceding fork. If the frequency of the last tuning fork is double than the first fork, then the frequency of the second tuning fork is \(\ldots \ldots\) (A) 132 (B) 138 (C) 276 (D) 144

A tuning fork arrangement produces 4 beats/second with one fork of frequency \(288 \mathrm{~Hz}\). A little wax is applied on the unknown fork and it then produces 2 beats/s. The frequency of the unknown fork is $\ldots \ldots \ldots . \mathrm{Hz}$. (A) 286 (B) 292 (C) 294 (D) 288

If the maximum frequency of a sound wave at room temperature is $20,000 \mathrm{~Hz}\( then its minimum wavelength will be approximately \)\ldots \ldots\left(\mathrm{v}=340 \mathrm{~ms}^{-1}\right)$ (A) \(0.2 \AA\) (B) \(5 \AA\) (C) \(5 \mathrm{~cm}\) to \(2 \mathrm{~m}\) (D) \(20 \mathrm{~mm}\)
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