Question

The frequency of oscillation of a particle of mass m suspended at the end of a vertical spring having a spring constant k is directly proportional to ____________.

पर्याय

• mk

• m/k

• m²k

• `(k/m)^(1//2)`

Answer 1

The frequency of oscillation of a particle of mass m suspended at the end of a vertical spring having a spring constant k is directly proportional to `(k/m)^(1//2)`.

Explanation:

`"F" = "ma" = -"kx"`    ....(i)

`"As x" = "A" sin omega"t",`

`therefore  "a" = -omega^2 sin omega"t" = -omega^2 "x"`

`"Substituting a" = -omega^2" x in equation (i), we get,"`

`-momega^2 "x" = -"kx"  "or"  omega^2 = "k"/"m"`

`"Now,"  omega = 2pi"f" = ("k"/"m")^(1//2)`

`therefore  "f" = 1/(2pi) = ("k"/"m")^(1//2)`

`therefore  F prop ("k"/"m")^(1//2)`