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Q. No.A particle executes 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. | \( \frac{T}{12} \) | 2. | \(\frac{5 T}{12} \) |
| 3. | \( \frac{7 T}{12} \) | 4. | \(\frac{2 T}{3}\) |
Subtopic: Â Linear SHM |
 57%
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The displacement of a particle varies according to the relation:
\(x=4\left({\cos{\pi t}+\sin{\pi t}}\right)\)
The amplitude of the particle is:
1. \(8\)
2. \(-4\)
3. \(4\)
4. \(4\sqrt{2}\)
\(x=4\left({\cos{\pi t}+\sin{\pi t}}\right)\)
The amplitude of the particle is:
1. \(8\)
2. \(-4\)
3. \(4\)
4. \(4\sqrt{2}\)
Subtopic: Â Linear SHM |
 68%
From NCERT
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A particle moves so that its acceleration \(a\) is given by; \(a=-bx,\) where \(x \) is the displacement from the equilibrium position and \(b\) is a constant. The period of oscillation is:
1. \({2}\mathit{\pi}\sqrt{b}\)
2. \(\dfrac{2\pi }{\sqrt{b}}\)
3. \(\dfrac{2\pi }{b}\)
4. \(2\sqrt{\dfrac{\pi }{b}}\)
1. \({2}\mathit{\pi}\sqrt{b}\)
2. \(\dfrac{2\pi }{\sqrt{b}}\)
3. \(\dfrac{2\pi }{b}\)
4. \(2\sqrt{\dfrac{\pi }{b}}\)
Subtopic: Â Linear SHM |
 86%
From NCERT
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A body oscillates with SHM according to the equation (in SI units), \(x= 5\cos\left[2\pi t +\frac{\pi}{4}\right].\) At \(t = 1.5\) s, acceleration of the body will be:
| 1. | \(140 \text{ cm} / \text{s}^2 \) | 2. | \(160 \text{ m} / \text{s}^2 \) |
| 3. | \(140 \text{ m} / \text{s}^2 \) | 4. | \(14 \text{ m} / \text{s}^2\) |
Subtopic: Â Linear SHM |
 60%
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A particle moves in the x-y plane according to the equation
\(x = A \cos^2 \omega t\) and \(y = A \sin^2 \omega t\)
Then, the particle undergoes:
| 1. | uniform motion along the line \(x + y = A\) |
| 2. | uniform circular motion along \(x^2 + y^2 = A^2\) |
| 3. | SHM along the line \(x + y = A\) |
| 4. | SHM along the circle \(x^2 + y^2 = A^2\) |
Subtopic: Â Linear SHM |
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A particle executes simple harmonic motion between \(x=-A\) and \(x=+A.\) The time taken for it to move from \(0\) to \(A/2\) is \(T_1\) and the time to move 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. \(T_{1}=2 T_{2}\)
1. \(T_{1}<T_{2}\)
2. \(T_{1}>T_{2}\)
3. \(T_{1}=T_{2}\)
4. \(T_{1}=2 T_{2}\)
Subtopic: Â Linear SHM |
 74%
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The figure given below depicts two circular motions. The radius of the circle, the period of revolution, the initial position, and the sense of revolution are indicated in the figures. Equations of the x-projection of the radius vector of the rotating particle \(\mathrm P\) in each case are, respectively:
| 1. | \({x}({t})={A} \cos \left(\dfrac{2 \pi}{4} {t}+\dfrac{\pi}{4}\right)\text{ and }{x}({t})={B} \cos \left(\dfrac{\pi}{15} {t}-\dfrac{\pi}{2}\right)\) |
| 2. | \({x}({t})={A} \cos \left(\dfrac{2 \pi}{4} {t}+\dfrac{\pi}{4}\right)\text{ and }{x}({t})={B} \sin \left(\dfrac{\pi}{15} {t}-\dfrac{\pi}{2}\right)\) |
| 3. | \({x}({t})={A} \cos \left(\dfrac{2 \pi}{4} {t}+\dfrac{\pi}{4}\right)\text{ and }{x}({t})={B} \cos \left(\dfrac{\pi}{15} {t}-\dfrac{\pi}{4}\right)\) |
| 4. | \({x}({t})={A} \sin \left(\dfrac{2 \pi}{4} {t}+\dfrac{\pi}{4}\right)\text{ and }{x}({t})={B} \cos \left(\dfrac{\pi}{15} {t}-\dfrac{\pi}{2}\right)\) |
Subtopic: Â Linear SHM |
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