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Simple Harmonic Oscillation

the amplitude...

Question

the amplitude of the simple harmonic motion is Y=(3 sin omega t+4 cos omega t)m

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Q. A simple harmonic motion is represented by

x = 7 s i n ω t +

c o s^{2} \frac{ω t}{2}

, then the amplitude of motion is approximately

Q. The resultant amplitude due to superposition of three simple harmonic motions

x_{1} = 3 sin ω t

x_{2} = 5 sin (ω t + 37^{\circ})

x_{3} = - 15 cos ω t

Q. The motion of a particle is given by x = A sin ωt + B cos ωt. The motion of the particle is
(a) not simple harmonic
(b) simple harmonic with amplitude A + B
(c) simple harmonic with amplitude (A + B)/2
(d) simple harmonic with amplitude

\sqrt{A^{2} + B^{2}} .

Q. Two simple harmonic motions are given by

x_{1} = \frac{a}{2} sin ω t + \frac{\sqrt{3} a}{2} cos ω t

x_{2} = a sin ω t + a cos ω t

. What is the ratio of the amplitude of the first motion to that of the second ?

Q. The instanteneous displacement of

x

of a particle executing simple harmonic motion is given by

x = a_{1} sin ω t + a_{2} cos (ω t + π / 6)

. The amplitude

A

of oscillation is given by

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