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Chapter 10: Problem 1358

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) \(3 \mathrm{~N}\)

Short Answer

Expert verified
The maximum force acting on the body is 0.6 N (Option A).

Step by step solution

01

Identify the given values and the formula to find maximum acceleration

We are given the mass (m) of the body, the amplitude (A) of the oscillation, and the periodic time (T). We will use the maximum acceleration (a_max) formula in SHM, which is: a_max = ω²A, where ω is the angular frequency, and it is related to the periodic time by the formula: ω = 2π/T.
02

Calculate the angular frequency (ω)

Using the formula ω = 2π/T, we can find ω. We are given T = π/10 s, so we have: ω = 2π/(π/10) ω = 2 × 10 ω = 20 rad/s
03

Calculate the maximum acceleration (a_max)

Now, we can use the maximum acceleration formula: a_max = ω²A. We have found ω and the given amplitude A = 0.3 m, so we can calculate a_max: a_max = (20 rad/s)² × 0.3 m a_max = 400 × 0.3 m a_max = 120 m/s²
04

Calculate the maximum force (F_max)

Finally, we will use Newton's second law of motion (F = ma) to find the maximum force acting on the body. We have the mass (m = 5 g) and maximum acceleration (a_max = 120 m/s²). First, we need to convert the mass from grams to kilograms: m = 5 g × (1 kg/1000 g) m = 0.005 kg Now, we can calculate the maximum force (F_max): F_max = m × a_max F_max = 0.005 kg × 120 m/s² F_max = 0.6 N So, the maximum force acting on the body is 0.6 N, and the correct option is (A).

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Most popular questions from this chapter

The function \(\sin ^{2}(\omega t)\) represents (A) A SHM with periodic time \(\pi / \omega\) (B) A SHM with a periodic time \(2 \pi / \omega\) (C) A periodic motion with periodic time \(\pi / \omega\) (D) A periodic motion with period \(2 \pi / \omega\)

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