A spring of unstretched length 40 cm and spring constant k is attached to a block of mass 1 kg to one of its end. The other end of the spring is fixed on the top of a frictionless inclined plane of inclination `theta = 45^(@)` as shown in the figure, so that the spring extends by 3 cm. When the mass is displaced slightly and released the time period of the resulting oscillation is T. Determine the value of 5T close to nearest integer.

Answer

2

Explanation

From the figure, we can say that the force which increases the length of the spring by x = 3 cm is `F = mg sin theta`. Therefore, the spring constant is
`k = (F)/(x) = (mg sin theta)/(x)`
Now time period `T = 2pisqrt((m)/(k)) = 2pi sqrt((m)/(k)) = 2pi sqrt((m)/(mg sin theta//x))`
` = 2pi sqrt((x)/(g sin theta))`
Putting `x = 3 cm = 3 xx 10^(-2) m, g = 9.8 ms^(-2)` and `theta = 45^(@)`, we get
`T = 2pisqrt((x)/(g sin theta)) = 2pi sqrt((3 xx 10^(-2))/(9.8 xx sin 45))`
` = 2pi sqrt((3 xx 10^(-2))/(9.8 xx 0.707)) = 0.4132s`