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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) \(b\) (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

A person standing in a stationary lift measures the periodic time of a simple pendulum inside the lift to be equal to \(\mathrm{T}\). Now, if the lift moves along the vertically upward direction with an acceleration of $(\mathrm{g} / 3)$, then the periodic time of the lift will now be \((\mathrm{A}) \sqrt{3} \mathrm{~T}\) (B) \(\sqrt{(3 / 2) \mathrm{T}}\) (C) \((\mathrm{T} / 3)\) (D) \((\mathrm{T} / \sqrt{3})\)

The periodic time of a S.H.O. oscillating about a fixed point is $2 \mathrm{~s}$. After what time will the kinetic energy of the oscillator become \(25 \%\) of its total energy? (A) \(1 / 12 \mathrm{~s}\) (B) \(1 / 6 \mathrm{~s}\) (C) \(1 / 4 \mathrm{~s}\) (D) \(1 / 3 \mathrm{~s}\).

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.

A system is executing S.H.M. with a periodic time of \(4 / 5 \mathrm{~s}\) under the influence of force \(\mathrm{F}_{1}\) When a force \(\mathrm{F}_{2}\) is applied, the periodic time is \((2 / 5) \mathrm{s}\). Now if \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2}\) are applied simultaneously along the same direction, the periodic time will be......... (A) \(\\{4 /(5 \sqrt{5})\\}\) (B) \(\\{4 /(4 \sqrt{5})\\}\) (C) \(\\{8 /(4 \sqrt{5})\\}\) (D) \(\\{8 /(5 \sqrt{5})\\}\)

A particle executing S.H.M. has an amplitude \(\mathrm{A}\) and periodic time \(\mathrm{T}\). The minimum time required by the particle to get displaced by \((\mathrm{A} / \sqrt{2})\) from its equilibrium position is $\ldots \ldots \ldots \mathrm{s}$. (A) \(\mathrm{T}\) (B) \(\mathrm{T} / 4\) (C) \(\mathrm{T} / 8\) (D) \(\mathrm{T} / 16\)

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