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प्रश्न
A mass 'M' is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes S.H.M. of period T. If the mass is increased by 'm', the time period becomes `("'5T'")/3`. What is the ratio `("M"/"m")`?
पर्याय
`25/9`
`9/16`
`16/9`
`9/25`
MCQ
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उत्तर
`9/16`
Explanation:
T = `2pisqrt("M"/"k"); "5T"/3 = 2pisqrt(("M + m")/"k")`
∴ `5/3 xx 2pisqrt("M"/"k") = 2pisqrt(("M + m")/"k")`
∴ `25/9 * "M"/"k" = ("M + m")/"k"`
∴ `25/9 "M" = "M + m"`
Dividing by M, `25/9 = 1 + "m"/"M"`
∴ `"m"/"M" = 25/9 - 1 = 16/9`
`"M"/"m" = 9/16`
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Damped Oscillations
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