In the arrangement shown in figure, pulleys are light and spring are ideal. `K_(1)`, `k_(2)`, `k_(3)`and `k_(4)` are force constant of the spring. Calculate period of small vertical oscillations of block of mass `m`.
Correct Answer – A::B::C::D
Let block be displacement by x than displacement in springs be
`x_(1),x_(2),x_(3)` and `x_(4)`
Such that `x = 2x_(1) + 2x_(2) + 2x_(3) + 2x_(4)`
Now let restoring force on m be `F = kx` then
`2f = k_(1)x_(1) = k_(2)x_(2) = k_(3)x_(3) = k_(4)x_(4)`
`rArr F/k = (4F)/(k_(1)) + (4F)/(k_(2)) + (4F)/(k_(3)) + (4F)/(k_(4))`
`rArr 1/k = 4(1/(k_(1)) + 1/(k_(2)) + 1/(k_(3) ) + 1/(k_(4)))`
`T = 2pisqrt((m)/(k)) = 2pisqrt(4m(1/(k_(1))+(1)/(k_(2)) + 1/(k_(3)) + 1/(k_(4))))`
Correct Answer – A::B::C::D
When the mass `m` is displaced from its mean position by a distance `x`, let `F` be the restoring (extra tension) force produced in the string. By this extra tension further elongation in the springs are
`(2F)/(k_(1))` , `(2F)/(k_(2))`, `(2F)/(k_(3))` and `(2F)/(k_(4))` respectively.
Then,
`x = 2((2F)/(k_(2))) + 2((2F)/(k_(2))) + 2((2F)/(k_(3))) + 2((2F)/(k_(4)))`
or `F((4)/(k_(1)) + (4)/(k_(2)) + (4)/(k_(3)) + (4)/(k_(4))) = – x`
Here negative sign shows the restoring nature of force.
`a = – (x)/(m((4)/(k_(1)) + (4)/(k_(2)) + (4)/(k_(3)) + (4)/(k_(4)))`
`T = 2pi sqrt |(x)/(a)|`
`= 4pi sqrt (m((1)/(k_(1)) + (1)/(k_(2)) + (1)/(k_(3)) + (1)/(k_(4))))`