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प्रश्न
The equation of S.H.M. of a particle of amplitude 4 cm performing 150 oscillations per minute starting with an initial phase 30° is ____________.
विकल्प
`"x" = 4 "sin" (2 pi"t"+ pi//6)`
`"x" = 4 "sin" (5 pi"t"+ pi//6)`
`"x" = 4 "sin" (2 pi"t"+ pi//3)`
`"x" = 4 "sin" (5 pi"t"+ pi//3)`
MCQ
रिक्त स्थान भरें
उत्तर
The equation of S.H.M. of a particle of amplitude 4 cm performing 150 oscillations per minute starting with an initial phase 30° is `"x" = 4 "sin" (5 pi"t"+ pi//6)`.
Explanation:
`"A" = 4"cm", alpha = 30° = pi/6`
f = 150 oscillation per minute= 150/60 Hz,
`omega = 2pi"f" = 2pi xx 150/60 = 5pi`
Equation of S.H.M becomes,
`"x" = 4 "sin" (5 pi"t"+ pi//6)`
shaalaa.com
Amplitude (A), Period (T) and Frequency (N) of S.H.M.
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