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Chapter 10: Problem 1425

The equation for displacement of a particle at time \(\mathrm{t}\) is given by the equation \(\mathrm{y}=3 \cos 2 \mathrm{t}+4 \sin 2 \mathrm{t}\). If the mass of the particle is \(5 \mathrm{gm}\), then the total energy of the particle is \(\ldots \ldots \ldots\) erg (A) 250 (B) 125 (C) 500 (D) 375

Short Answer

Expert verified
Total energy of the particle is given by the expression: \[Total\:Energy = \frac{1}{2}(0.005)((-6\sin(2t) + 8\cos(2t))^2) + (0.005)(9.81)(3\cos(2t) + 4\sin(2t))\]. None of the options (A, B, C, D) are correct, as the total energy is not a constant value but a function of time t.

Step by step solution

01

Find the velocity of the particle

To find the velocity, we need to differentiate the displacement equation with respect to time: \[\frac{dy}{dt} = \frac{d}{dt}(3\cos(2t) + 4\sin(2t))\] Apply the chain rule and differentiate each term: \[\frac{dy}{dt} = -6\sin(2t) + 8\cos(2t)\] This is the velocity of the particle.
02

Calculate the kinetic energy

Now, we need to calculate the kinetic energy using the formula: \[KE = \frac{1}{2}mv^2\] First, we have the mass in grams, so we need to convert it to kg: \[m = 5gm = 0.005kg\] Now, square the velocity: \[\left(\frac{dy}{dt}\right)^{2} = (-6\sin(2t) + 8\cos(2t))^2\] Substitute the mass and velocity into the kinetic energy formula: \[KE = \frac{1}{2}(0.005)((-6\sin(2t) + 8\cos(2t))^2)\]
03

Calculate the potential energy

To calculate the potential energy, first, find the height, which in this case is the displacement y: \[h = y = 3\cos(2t) + 4\sin(2t)\] Now, substitute the mass and height into the potential energy formula: \[PE = (0.005)(9.81)(3\cos(2t) + 4\sin(2t))\]
04

Calculate the total energy

Finally, add the kinetic and potential energies to find the total energy of the particle: \[Total\:Energy = KE + PE\] \[Total\:Energy = \frac{1}{2}(0.005)((-6\sin(2t) + 8\cos(2t))^2) + (0.005)(9.81)(3\cos(2t) + 4\sin(2t))\] The total energy is not a constant value as it is a function of t. So, none of the answer options (A, B, C, D) are correct since they all represent constant values.

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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) \(b\) (C) \(\mathrm{c}\) (D) d Statement \(-1:\) For a particle executing S.H.M. with an amplitude of $0.01 \mathrm{~m}\( and frequency \)30 \mathrm{hz}\(, the maximum acceleration is \)36 \pi^{2} \mathrm{~m} / \mathrm{s}^{2}$. Statement \(-2:\) The maximum acceleration for the above particle is $\pm \omega 2 \mathrm{~A}\(, where \)\mathrm{A}$ is amplitude. (A) a (B) \(\mathrm{b}\) (C) \(c\) (D) \(\mathrm{d}\)

If two almost identical waves having frequencies \(\mathrm{n}_{1}\) and \(\mathrm{n}_{2}\), produced one after the other superposes then the time interval to obtain a beat of maximum intensity is \(\ldots \ldots \ldots .\) (A) \(\left\\{1 /\left(\mathrm{n}_{1}-\mathrm{n}_{2}\right)\right\\}\) (B) \(\left(1 / \mathrm{n}_{1}\right)-\left(1 / \mathrm{n}_{2}\right)\) (C) \(\left(1 / \mathrm{n}_{1}\right)+\left(1 / \mathrm{n}_{2}\right)\) (D) \(\left\\{1 /\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right)\right\\}\)

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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) \(\mathrm{d}\)

A string of linear density \(0.2 \mathrm{~kg} / \mathrm{m}\) is stretched with a force of \(500 \mathrm{~N}\). A transverse wave of length \(4.0 \mathrm{~m}\) and amplitude \(1 / 1\) meter is travelling along the string. The speed of the wave is \(\ldots \ldots \ldots \ldots \mathrm{m} / \mathrm{s}\) (A) 50 (B) \(62.5\) (C) 2500 (D) \(12.5\)
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