Two identical balls (A) and (B) each of mass (0.1 kg), are attached to two identical massless springs. The spring - mass system is constrained to move inside a riged smooth pipe bant in the form of a circle as shown in Fig. The pipe is fixed in a horizontal plane.
The centres of the balls can move in a circle of radius (0.06 pi) meter. Each spring has a natural length `zof0.06 pi` meter and spring spring constant5 `0.1 N//m`. Initially, both the balls are displaced by an angle `theta = pi// 6` radian with respect to the diameter (pQ) of the circle (as shown in Fig.) and released from, rest.
.
(i) Calculate the frequency of oscillation of ball (B).
(ii) Find the speed of ball (A) when (A) and (B) are at the two ends of the diameter (PQ).
(iii) What is the total energy of the system.

Options

A. `3.9xx10^(-4)J`
B. `10^(-4)J`
C. `6xx10^(-4)J`
D. `2xx10^(-4)J`

Answer

A

Explanation

Given-Mass of each block `A` and `b, m = 0.1kg` Radius of circle, `R = 0.06m`

Natural length of spring `l_(0) = 0.06 pi = piR` (Half circle) and spring constant, `k = 0.1N//m` In the streteched position elongation in each spring `x =R theta`. Let us draw `FBD` of `A`

Spring in lower side is stretched by `2x` and on upper side compressed by `2x`. Therefore, each spring will exert a force `2kx` on each block. Hence, a restroing force, `F =5kx` will acr on `A` in the direction shown in figure.
`tai =- F.R =-(4kx)R =- (4kR theta) R or tau =- 4kR^(2). theta(1)`
Since, each ball executes angular `SHM` about origin `O`. Eq. (1) can be rewritten as
`I alpha =- 4 kR^(2) q or (mR^(2))alpha =- 4kR^(2) theta or alpha =- ((4k)/(m)) theta`
`:.` Frequency oscillation, `f = (1)/(2pi) sqrt(|("acceleration")/("displacement")|)=(1)/(2pi)sqrt(|(alpha)/(theta)|)`
`f = (1)/(2pi) sqrt((4k)/(m))` Substituting the values, we hae
`f = (1)/(2pi) sqrt((4xx0.1)/(0.1)) =(1)/(pi) Hz`
(ii) In stretched position, potential energy of the system is `PE =2 {(1)/(2)k} {2x}^(2) =4 kx^(2)` and in mean position, both the blocks have kinetic enegry only.
Hence, `KE = 2 {(1)/(2)mv^(2)} =mv^(2)` From energy conservation
`PE = KE, :. 4 kx^(2) =mv^(2)`
`:. v = 2x sqrt((k)/(m)) = 2R theta sqrt((k)/(m))`
`:. v = 2(0.06) ((pi)/(6)) sqrt((0.1)/(0.1)) or v = 0.0628 m//s`
(iii) Total energy of the sysetm, `E = PE` in stretched position or `=KE` in mean position
`E = mv^(2) =(0.1) (0.0628)^(2) J or E = 3.9 xx 10^(-4)J`