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Chapter 10: Problem 1385

A body of mass \(1 \mathrm{~kg}\) suspended from the free end of a spring having force constant \(400 \mathrm{Nm}^{-1}\) is executing S.H.M. When the total energy of the system is 2 joule, the maximum acceleration is $\ldots \ldots . \mathrm{ms}^{-2}$. (A) \(8 \mathrm{~ms}^{-2}\) (B) \(10 \mathrm{~ms}^{-2}\) (C) \(40 \mathrm{~ms}^{-2}\) (D) \(40 \mathrm{cms}^{-2}\)

Short Answer

Expert verified
The maximum acceleration of the body is 40 m/s², which corresponds to option (C).

Step by step solution

01

Identify the given parameters

We are given the following parameters: Mass of the body (m) = 1 kg Force constant of the spring (k) = 400 Nm^{-1} Total mechanical energy of the system (E) = 2 J We are asked to find the maximum acceleration of the body.
02

Understanding energy conservation in SHM

In simple harmonic motion, the total mechanical energy of the system remains constant and is the sum of the kinetic energy (KE) and the potential energy (PE) stored in the spring. At the maximum displacement (amplitude, A) of the oscillations, the kinetic energy is zero, and all energy is in the form of potential energy stored in the spring. So, the total energy at maximum displacement can be expressed as: E = PE = \(\frac{1}{2}kA^2\)
03

Finding the amplitude

We have the total mechanical energy, and we can solve for the amplitude (A) using the conservation of mechanical energy: E = \(\frac{1}{2}kA^2\) 2 = \(\frac{1}{2}(400)A^2\) To find the amplitude, we can now solve for A: A^2 = \(\frac{2}{200}\) A = \(\sqrt{\frac{1}{100}}\) = 1/10 m
04

Finding the maximum acceleration

We can now find the maximum acceleration using the relationship between acceleration and displacement in simple harmonic motion: a = -k * x / m At the maximum displacement (x = A), the acceleration will be at its maximum value. Therefore: a_max = -k * A / m a_max = -400 * (1/10) / 1 a_max = -40 m/s^2 The maximum acceleration is negative because it acts in the opposite direction to the displacement. However, the magnitude of the maximum acceleration is: |a_max| = 40 m/s^2 So, the maximum acceleration of the body is 40 m/s², which corresponds to option (C).

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Most popular questions from this chapter

The distance travelled by a particle performing S.H.M. during time interval equal to its periodic time is \(\ldots \ldots\) (A) A (B) \(2 \mathrm{~A}\) (C) \(4 \mathrm{~A}\) (D) Zero.

When two sound waves having amplitude A, angular frequency \(\omega\) and a phase difference of \(\pi / 2\) superposes, the maximum amplitude and angular frequency of the resultant wave is \(\ldots \ldots \ldots \ldots\) (A) \(\sqrt{2} \mathrm{~A}, \omega\) (B) \((\mathrm{A} / \sqrt{2}),(\omega / 2)\) (C) \((\mathrm{A} / \sqrt{2}), \omega\) (D) \(\sqrt{2} \mathrm{~A},(\omega / 2)\)

Two sound waves are represented by $\mathrm{y}=\mathrm{a} \sin (\omega \mathrm{t}-\mathrm{kx})\( and \)\mathrm{y}=\mathrm{a} \cos (\omega \mathrm{t}-\mathrm{kx})$. The phase difference between the waves in water is \(\ldots \ldots \ldots\) (A) \((\pi / 2)\) (B) \((\pi / 4)\) (C) \(\pi\) (D) \((3 \pi / 4)\)

A small spherical steel ball is placed at a distance slightly away from the center of a concave mirror having radius of curvature \(250 \mathrm{~cm}\). If the ball is released, it will now move on the curved surface. What will be the periodic time of this motion? Ignore frictional force and take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$. (A) \((\pi / 4) \mathrm{s}\) (B) \(\pi \mathrm{s}\) (C) \((\pi / 2) \mathrm{s}\) (D) \(2 \pi \mathrm{s}\)

As shown in figure, a block A having mass \(M\) is attached to one end of a massless spring. The block is on a frictionless horizontal surface and the free end of the spring is attached to a wall. Another block B having mass ' \(\mathrm{m}\) ' is placed on top of block A. Now on displacing this system horizontally and released, it executes S.H.M. What should be the maximum amplitude of oscillation so that B does not slide off A? Coefficient of static friction between the surfaces of the block's is \(\mu\). (A) \(A_{\max }=\\{(\mu \mathrm{mg}) / \mathrm{k}\\}\) (B) \(A_{\max }=[\\{\mu(m+M) g\\} / k]\) (C) \(A_{\max }=[\\{\mu(M-\mathrm{m}) g\\} / \mathrm{k}]\) (D) \(A_{\max }=[\\{2 \mu(M+m)\\} / k]\)
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