Q. No.A particle is executing SHM with an amplitude \(A\) and the time period \(T\). If at \(t=0\), the particle is at its origin (mean position), then the time instant when it covers a distance equal to \(2.5A\) will be:
| 1. | \( \dfrac{T}{12} \) | 2. | \(\dfrac{5 T}{12} \) |
| 3. | \( \dfrac{7 T}{12} \) | 4. | \(\dfrac{2 T}{3}\) |
Subtopic: Â Linear SHM |
 56%
From NCERT
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A particle is subjected to two simple harmonic motions in the same direction having equal amplitudes and equal frequency. If the resulting amplitude is equal to the amplitude of individual motions, the phase difference between them will be:
1. \(\frac{\pi}{3}\)
2. \(\frac{2\pi}{3}\)
3. \(\frac{\pi}{6}\)
4. \(\frac{\pi}{2}\)
1. \(\frac{\pi}{3}\)
2. \(\frac{2\pi}{3}\)
3. \(\frac{\pi}{6}\)
4. \(\frac{\pi}{2}\)
Subtopic: Â Linear SHM |
 61%
From NCERT
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The equation of an SHM is given as \(y = 3\sin\omega t+ 4\cos \omega t\) where \(y\) is in centimeters. The amplitude of the SHM will be?
| 1. | \(3~\text{cm}\) | 2. | \(3.5~\text{cm}\) |
| 3. | \(4~\text{cm}\) | 4. | \(5~\text{cm}\) |
Subtopic: Â Linear SHM |
 90%
From NCERT
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A particle executes linear SHM between \(x=A.\) The time taken to go from \(0\) to \(A/2\) is \(T_1\) and to go from \(A/2\) to \(A\) is \(T_2\) then:
| 1. | \(T_1<T_2\) | 2. | \(T_1>T_2\) |
| 3. | \(T_1=T_2\) |
4. | \(T_1= 2T_2\) |
Subtopic: Â Linear SHM |
 71%
From NCERT
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A particle of mass \(m\) and charge \(\text-q\) moves diametrically through a uniformly charged sphere of radius \(R\) with total charge \(Q\). The angular frequency of the particle's simple harmonic motion, if its amplitude \(<R\), is given by:
1. \(\sqrt{\dfrac{qQ}{4 \pi \varepsilon_0 ~mR} }\)
2. \(\sqrt{\dfrac{qQ}{4 \pi \varepsilon_0 ~mR^2} }\)
3. \(\sqrt{\dfrac{qQ}{4 \pi \varepsilon_0 ~mR^3}}\)
4. \( \sqrt{\dfrac{m}{4 \pi \varepsilon_0 ~qQ} }\)
Subtopic: Â Linear SHM |
 57%
From NCERT
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If a particle is executing SHM, with an amplitude \(A\), the distance moved and the displacement of the body in a time equal to its time period are, respectively:
| 1. | \(2A,A\) | 2. | \(4A,0\) |
| 3. | \(A,A\) | 4. | \(0,2A\) |
Subtopic: Â Linear SHM |
 83%
From NCERT
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A particle moves according to the law \(x= r\cos\left(\frac{\pi t}{2}\right)\). The distance covered by it in the time interval between \(t=0\) to \(t=3~\text{s}\) will be:
| 1. | \(r\) | 2. | \(2r\) |
| 3. | \(3r\) | 4. | \(4r\) |
Subtopic: Â Linear SHM |
From NCERT
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A particle is executing linear simple harmonic motion with an amplitude \(a\) and an angular frequency \(\omega\). Its average speed for its motion from extreme to mean position will be:
1. \(\frac{a\omega}{4}\)
2. \(\frac{a\omega}{2\pi}\)
3. \(\frac{2a\omega}{\pi}\)
4. \(\frac{a\omega}{\sqrt{3}\pi}\)
Subtopic: Â Linear SHM |
 56%
From NCERT
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Two particles are oscillating along two close parallel straight lines side by side, with the same frequency and amplitudes. They pass each other, moving in opposite directions when their displacement is half of the amplitude. The mean positions of the two particles lie in a straight line perpendicular to the paths of the two particles. The phase difference is:
1. \(\frac{\pi}{6}\)
2. \(0\)
3. \(\frac{2\pi}{3}\)
4. \(\pi\)
1. \(\frac{\pi}{6}\)
2. \(0\)
3. \(\frac{2\pi}{3}\)
4. \(\pi\)
Subtopic: Â Linear SHM |
 56%
From NCERT
AIPMT - 2011
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The amplitude and the time period in an SHM are \(0.5\) cm and \(0.4\) sec respectively. If the initial phase is \(\frac{\pi}{2}\) radian, then the equation of SHM will be:
1. \(y = 0.5\sin(5\pi t)\)
2. \(y = 0.5\sin(4\pi t)\)
3. \(y = 0.5\sin(2.5\pi t)\)
4. \(y = 0.5\cos(5\pi t)\)
1. \(y = 0.5\sin(5\pi t)\)
2. \(y = 0.5\sin(4\pi t)\)
3. \(y = 0.5\sin(2.5\pi t)\)
4. \(y = 0.5\cos(5\pi t)\)
Subtopic: Â Linear SHM |
 69%
From NCERT
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