InterviewSolution
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InterviewSolution
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Two identical balls `A and B` eavh of mass `0.1 kg` are attached to two identical mass less is springs. The spring-mass system is consetrained to move inside a right smooth pipe bent in the form of a circle as shown in figure . The pipe is fixed in a horigental plane. The center of the balls can move in a circle of radius `0.06 m`. Each spring has a natural length of `0.06 pi m` and force constant `0.1 N//m`. Initially, both the balls are displaced bu an angle `theta = pi//6` dadius withrespect to diameter PQ of the circle and released from rest. a. Calculate the frequency of oscillation of the ball B. b. What is the total energy of the system ? c. Find the speed of the bal A when A and B are at the two ends of the diameter PQ. |
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Answer» a.As here two masses `A and B` are connected by two sprins , this problem is equivelent to the oscillation of a reduced mass `m` by a spring of effective force constant `k_(eff)` given by `m = (m_(1) m_(2))/(m_(1) + m_(2)) = (0.1 xx 0.1)/(0.1 + 0.1) = 0.5 kg` and `k_(eff) = k_(1) + k_(2) = 0.1 + 0.1 = 0.2 N/m` So `f = (1)/(2 pi) sqrt((k_(eq)/(m)) = (1)/(2 pi) sqrt((0.20)/(0.05)) = (1)/(pi) Hz` b. As here one spring is compresent while the other is atreched by same amount (say y) and balls are at rest at `A and B`, so `E = (1)/(2) k_(1) y^(2) + (1)/(2)k_(2) y^(2) = k y^(2)` `[as k_(1) = k_(2)]` But from above figure `y = y_(1) + y_(2) = R theta_(1) + R theta_(2) = 2 R theta` `y = 2 xx 0.06 xx (pi//6) = 0.02 pi m` So `E = (0.1) (0.02 theta)^(2) = 4 pi^(2) xx 10^(-5) J` c. As at P and Q springs are ustertched so the whole energy becomes kinetic of the balls A and B, i.e. `(1)/(2)m_(1) v_(1)^(2) + (1)/(2)m_(2) v_(2)^(2) = E = 4 pi^(2) xx 10^(-5) J` Here `m_(1) = m_(2) = 0.1 kg` and `v_(1) = v_(2) = v` So, ` 0.1 v^(2) = 4 pi^(2) xx 10^(-5), i.e. v = 2 pi xx 10^(-2) m//s` |
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