A vertical spring mass system has the same time period as simple pendulum undergoing small oscillations. Now, both of them are put in an elevator going downwards with an acceleration `5 m//s^(2)`. The ratio of time period of the spring mass system to the time period of the pendulum is (Assume, acceleration due to gravity, `g=10 m//s^(2)`)

Options

A. `sqrt(3/2)`
B. `sqrt(2/3)`
C. `1/sqrt(2)`
D. `sqrt(2)`

Answer

C

Explanation

We know that time period of a spring mass system,
`T_(i)=2 pi sqrt(m/k)`
where, m = mass of body
and k = force constant of the spring
Time period of simple pendulum,
`T_(2)=2pi sqrt(l/g)`
According to the question, initially time period of a spring mass system, `T_(1)=` time period of simple pendulum, `T_(2)`
`2pi sqrt(m/k)=2 pi sqrt(l/g)`
`sqrt(m/k)=sqrt(l/10)` ...(i) `[ :' g=10 m//s^(2)]`
When both of them are put in an elevator going downwards with an acceleration `5 m//s^(2)`, then no affect occurs on the time period of spring mass system.
i.e., `T_(1)'=T_(1)=2pi sqrt(m/k)`
But time period of simple pendulum changes in elevator and is given by
`T_(2)'=2pi sqrt(l/(g_("eff")))`, where `g_("eff")=g-a=` effective acceleration of the pendulum when elevator is accelerating downwards with a `m//s^(2)`
`T_(2)'=2pi sqrt(l/(g-5))=2pi sqrt(l/5)`
`:. (T_(1)')/(T_(2)')=(2pi sqrt(m/k))/(2pi sqrt(l/5))=sqrt(l//10)/sqrt(l//5)` [From Eq. (i)]
`:. (T_(1)')/(T_(2)')=1/sqrt(2)`
Hence, the ratio of time period of the spring mass system to the time period of the pendulum is `1/sqrt(2)`.