Question

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")`?

विकल्प

• `25/9`

• `9/16`

• `16/9`

• `9/25`

Answer 1

`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`