`A` mass `M` is suspended from a spring of negligible mass. The spring is pulled a little then released, so that the mass executes simple harmonic motion of time period `T`. If the mass is increased by `m`, the time period becomes `(5T)/(3)`. Find the ratio of `m//M`.

`A` mass `M` is suspended from a spring of negligible mass. The spring

1.	`A` mass `M` is suspended from a spring of negligible mass. The spring is pulled a little then released, so that the mass executes simple harmonic motion of time period `T`. If the mass is increased by `m`, the time period becomes `(5T)/(3)`. Find the ratio of `m//M`.
Answer» Correct Answer - A `T = 2pi sqrt((M)/(k))` `(5T)/(3) = 2pi sqrt ((M + m)/(k))` Solving these two equation, we get `(m)/(M) = (16)/(9)`

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`A` mass `M` is suspended from a spring of negligible mass. The spring is pulled a little then released, so that the mass executes simple harmonic motion of time period `T`. If the mass is increased by `m`, the time period becomes `(5T)/(3)`. Find the ratio of `m//M`.

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