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) \(\mathrm{d}\)

Short Answer

Expert verified
The correct answer is (A) Both statement 1 and statement 2 are true, and statement 2 is the correct explanation of statement 1.

Step by step solution

01

Calculate the maximum acceleration for statement 1

Given the amplitude A = 0.01 m and frequency f = 30 Hz, we need to find the maximum acceleration. Since the maximum acceleration is given by the formula: \[a_{max} = \omega^2 A\] where \(\omega = 2\pi f\) is the angular frequency, we can calculate the maximum acceleration using the given values. Step 2: Calculate the angular frequency
02

Calculate the angular frequency

With f = 30 Hz, we can calculate the angular frequency as follows: \[\omega = 2\pi f = 2\pi(30) = 60\pi\ rad/s\] Step 3: Calculate the maximum acceleration
03

Calculate the maximum acceleration using the angular frequency

Now we can calculate the maximum acceleration: \[a_{max} = \omega^2 A = (60\pi)^2 (0.01 \mathrm{m}) = 36\pi^2\ \mathrm{m/s^2}\] Step 4: Analyze statement 2
04

Check the validity of the statement 2

We already calculated the maximum acceleration for the given particle using the formula \(a_{max} = \omega^2 A\). As Statement 2 suggests, this is indeed the correct formula. Therefore, statement 2 is true. Step 5: Comparison of statements and options
05

Compare statements and options

Both statement 1 and statement 2 are true. Additionally, statement 2 is indeed the correct explanation for statement 1. Therefore, the correct option is: (a) Statement 1 is true and statement 2 is true. Statement 2 is the correct explanation of statement 1.

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Most popular questions from this chapter

The ratio of frequencies of two waves travelling through the same medium is \(2: 5 .\) The ratio of their wavelengths will be.... (A) \(2: 5\) (B) \(5: 2\) (C) \(3: 5\) (D) \(5: 3\)

A particle is executing S.H.M. between \(\mathrm{x}=-\mathrm{A}\) and \(\mathrm{x}=+\mathrm{A}\). If the time taken by the particle to travel from \(\mathrm{x}=0\) to \(\mathrm{A} / 2\) is \(\mathrm{T}_{1}\) and that taken to travel from \(\mathrm{x}=\mathrm{A} / 2\) to \(\mathrm{x}=\mathrm{A}\) is \(\mathrm{T}_{2}=\) then \(\ldots .\) (A) \(\mathrm{T}_{1}<\mathrm{T}_{2}\) (B) \(\mathrm{T}_{1}>\mathrm{T}_{2}\) (C) \(\mathrm{T}_{1}=2 \mathrm{~T}_{2}\) (D) \(\mathrm{T}_{1}=\mathrm{T}_{2}\)

Equation for a harmonic progressive wave is given by \(\mathrm{y}=\mathrm{A}\) \(\sin (15 \pi t+10 \pi x+\pi / 3)\) where \(x\) is in meter and \(t\) is in seconds. This wave is \(\ldots \ldots\) (A) Travelling along the positive \(\mathrm{x}\) direction with a speed of $1.5 \mathrm{~ms}^{-1}$ (B) Travelling along the negative \(\mathrm{x}\) direction with a speed of $1.5 \mathrm{~ms}^{-1} .$ (C) Has a wavelength of \(1.5 \mathrm{~m}\) along the \(-\mathrm{x}\) direction. (D) Has a wavelength of \(1.5 \mathrm{~m}\) along the positive \(\mathrm{x}\) - direction.

The ratio of force constants of two springs is \(1: 5\). The equal mass suspended at the free ends of both springs are performing S.H.M. If the maximum acceleration for both springs are equal, the ratio of amplitudes for both springs is \(\ldots \ldots\) (A) \((1 / \sqrt{5})\) (B) \((1 / 5)\) (C) \((5 / 1)\) (D) \((\sqrt{5} / 1)\)

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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