Question

A simple harmonic motion is represented by y = 8 sin 10 pi t+ 5 cos 10 pi t, where y is in cm and t is in sec. Calculate the amplitude, time period and initial phase.

Answer 1

Here,

$y=8sin\,10\pi t + 5cos\,10\pi t\,\,\,...(i)$

The general equation of S.H.S is,

$y=a sin(\omega t+\delta)= a sin \times \omegat \times cos\delta + a cos \times \omega \times sin\delta\\\implies y = (acos\delta)sin\,\omega t+(asin\delta)cos\delta t \,\,\,\,...(ii)$

$Compairing\,\,(i)\,\,and \,\,(ii)\,\,we\,\,get,$

$a(cos\,\delta) = 8\,\,\,...(iii)\\a(sin\,\delta) = 5\,\,\,...(iv)$

$\omega t = 10\pi t \\\implies \omega = 10\pi\\\therefore Time\,\, Period, T = \frac{2\pi}{\omega} = \frac{2\pi}{10\pi}= 0.2s\\\\Squaring\,\,and\,\,adding\,\,(iii)\,\,and\,\,(iv)\,\,we \,\,get,\\a^{2}(cos^{2} \phi +sin^{2} \phi)=8^{2}+5^{2} = 89\\\implies a^{2}= 89 \,\,or\,\,a\approx9.43 \, cm\,\\\therefore Amplitude, a = 9.43\,cm\\\\Dividing\,\, (iv)\,\, by \,\,(iii) \,\,we \,\,get,\\tan\,\delta= \frac{5}{8}= 0.625\\\therefore \delta = 32^{\circ}\,(approx.)$

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Answer 2

Answer:

The minimum time after which the particle keeps on repeating its motion is known as the time period (or) the shortest time taken to complete one oscillation is also defined as the time period.

T = 2π/ω

Frequency: The number of oscillations per second is defined as the frequency.

Frequency = 1/T and, angular frequency ω = 2πf = 2π/T