The equation of an SHM is given as y=3sinωt + 4cosωt where y is in centimeters. The amplitude of the SHM will be?

 1 3 cm 2 3.5 cm 3 4 cm 4 5 cm
Subtopic:  Linear SHM |
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A particle executing simple harmonic motion of amplitude $$5~\text{cm}$$ has a maximum speed of $$31.4~\text{cm/s}.$$ The frequency of its oscillation will be:
1. $$1~\text{Hz}$$
2. $$3~\text{Hz}$$
3. $$2~\text{Hz}$$
4. $$4~\text{Hz}$$

Subtopic:  Linear SHM |
86%
From NCERT
AIPMT - 2005
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Two simple harmonic motions of angular frequency 100 rad s -1 and 1000 rad ${s}^{-1}$ have the same displacement amplitude. The ratio of their maximum acceleration will be:
1. 1:10
2. 1:102
3. 1:103
4. 1:104

Subtopic:  Linear SHM |
86%
From NCERT
NEET - 2008
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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:  Linear SHM |
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A body performs simple harmonic motion about x=0 with an amplitude a and a time period T. The speed of the body at $\mathrm{x}=\frac{\mathrm{a}}{2}$ will be:
1. $\frac{\mathrm{\pi a}\sqrt{3}}{2\mathrm{T}}$
2. $\frac{\mathrm{\pi a}}{\mathrm{T}}$
3. $\frac{3{\mathrm{\pi }}^{2}\mathrm{a}}{\mathrm{T}}$
4. $\frac{\mathrm{\pi a}\sqrt{3}}{\mathrm{T}}$

Subtopic:  Linear SHM |
77%
From NCERT
NEET - 2009
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The amplitude of a simple harmonic oscillator is $$A$$ and speed at the mean position is $$v_0$$ .The speed of the oscillator at the position $$x={A \over \sqrt{3}}$$ will be:

 1 $$2v_0 \over \sqrt{3}$$ 2 $$\sqrt{2}v_0 \over 3$$ 3 $${2 \over 3}v_0$$ 4 $$\sqrt{2}v_0 \over \sqrt{3}$$
Subtopic:  Linear SHM |
76%
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A particle executes SHM with a time period of 4 s. The time taken by the particle to go directly from its mean position to half of its amplitude will be:

1.  $\frac{1}{3}$ s

2.  1 s

3.  $\frac{1}{2}$ s

4.   2 s

Subtopic:  Linear SHM |
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A point performs simple harmonic oscillation of period $$\mathrm{T}$$ and the equation of motion is given by; $$x=a \sin (\omega t+\pi / 6)$$After the elapse of what fraction of the time period, the velocity of the point will be equal to half of its maximum velocity?
1. $$\frac{T}{8}$$

2. $$\frac{T}{6}$$

3. $$\frac{T}{3}$$

4. $$\frac{T}{12}$$

Subtopic:  Linear SHM |
70%
From NCERT
AIPMT - 2008
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A particle executes linear SHM between $$x=A.$$ The time taken to go from $$0$$ to $$A/2$$ is ${\mathrm{T}}_{1}$ and to go from $$A/2$$ to $$A$$ is ${\mathrm{T}}_{2}$, then:

 1 ${\mathrm{T}}_{1}$ $<$ 2 ${\mathrm{T}}_{1}$ $>$ ${\mathrm{T}}_{2}$ 3 ${\mathrm{T}}_{1}$ $=$ ${\mathrm{T}}_{2}$ 4 ${\mathrm{T}}_{1}$ $=$ $2{\mathrm{T}}_{2}$
Subtopic:  Linear SHM |
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The amplitude and the time period in an S.H.M. are 0.5 cm and 0.4 sec respectively. If the initial phase is $\mathrm{\pi }/2$ radian, then the equation of S.H.M. will be:

1. $\mathrm{y}=0.5\mathrm{sin}5\mathrm{\pi t}$

2. $\mathrm{y}=0.5\mathrm{sin}4\mathrm{\pi t}$

3. $\mathrm{y}=0.5\mathrm{sin}2.5\mathrm{\pi t}$

4. $\mathrm{y}=0.5\mathrm{cos}5\mathrm{\pi t}$

Subtopic:  Linear SHM |
69%
From NCERT
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