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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

For the following questions, statement as well as the reason(s) are given. Each questions has four options. Select the correct option. (a) Statement \(-1\) is true, statement \(-2\) is true; statement \(-2\) is the correct explanation of statement \(-1\). (b) Statement \(-1\) is true, statement \(-2\) is true but statement \(-2\) is not the correct explanation of statement \(-1\). (c) Statement \(-1\) is true, statement \(-2\) is false (d) Statement \(-1\) is false, statement \(-2\) is true (A) a (B) \(\mathrm{b}\) (C) \(c\) (D) \(\mathrm{d}\) Statement \(-1:\) For a particle executing SHM, the amplitude and phase is decided by its initial position and initial velocity. Statement \(-2:\) In a SHM, the amplitude and phase is dependent on the restoring force. (A) a (B) \(b\) (C) \(\mathrm{c}\) (D) \(\mathrm{d}\)

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) \((\pi / 3)\)

The equation for displacement of a particle at time \(t\) is given by the equation \(\mathrm{y}=3 \cos 2 \mathrm{t}+4 \sin 2 \mathrm{t}\). The periodic time of oscillation is \(\ldots \ldots \ldots \ldots\) (A) \(2 \mathrm{~s}\) (B) \(\pi \mathrm{s}\) (C) \((\pi / 2) \mathrm{s}\) (D) \(2 \pi \mathrm{s}\)

If the equation for displacement of two particles executing S.H.M. is given by \(\mathrm{y}_{1}=2 \sin (10 \mathrm{t}+\theta)\) and $\mathrm{y}_{2}=3 \cos 10 \mathrm{t}$ respectively, then the phase difference between the velocity of two particles will be \(\ldots \ldots \ldots\) (A) \(-\theta\) (B) \(\theta\) (C) \(\theta-(\pi / 2)\) (D) \(\theta+(\pi / 2)\).

If the resultant of two waves having amplitude \(\mathrm{b}\) is \(\mathrm{b}\), then the phase difference between the two waves is (A) \(120^{\circ}\) (B) \(60^{\circ}\) (C) \(90^{\circ}\) (D) \(180^{\circ}\)
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