Simple Harmonic Motion 4 Question 15

15. Two identical balls A and B, each of mass 0.1kg, are attached to two identical massless springs. The spring-mass system is constrained to move inside a rigid smooth pipe bent in the form of a circle as shown in figure. The pipe is fixed in a horizontal plane. The centres of the balls can move in a circle of radius 0.06m. Each spring has a natural length of 0.06πm and spring constant 0.1N/m. Initially, both the balls are displaced by an angle θ=π/6rad with respect to the diameter PQ of the circle (as shown in figure) and released from rest. (1993,6M)

(a) Calculate the frequency of oscillation of ball B.

(b) Find the speed of ball A when A and B are at the two ends of the diameter PQ

(c) What is the total energy of the system?

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Answer:

Correct Answer: 15. (a) 1πHz

(b) 0.0628m/s

(c) 3.9×104J

Solution:

  1. Given, mass of each block A and B,m=0.1kg

Radius of circle, R=0.06m

Natural length of spring l0=0.06π=πR (Half circle) and spring constant, k=0.1N/m

In the stretched position elongation in each spring

x=Rθ

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 restoring force, F=4kx will act on A in the direction shown in figure below .

Restoring torque of this force about origin

τ=FR=(4kx)R=(4kRθ)R or τ=4kR2θ

Since, τθ, each ball executes angular SHM about origin O.

Eq. (i) can be rewritten as

Iα=4kR2θ or (mR2)α=4kR2θ or =4kmθ

(a) Frequency of oscillation,

f=12π acceleration  displacement =12παθf=12π4km

Substituting the values, we have

f=12π4×0.10.1=1πHz

(b) In stretched position, potential energy of the system is

PE=212k2x2=4kx2

and in mean position, both the blocks have kinetic energy only. Hence, KE=212mv2=mv2

From energy conservation PE=KE

4kx2=mv2v=2xkm=2Rθkm

Substituting the values v=2(0.06)(π/6)0.10.1

 or v=0.0628m/s

(c) Total energy of the system, E

=PE in stretched position 

or =KE in mean position

E=mv2=(0.1)(0.0628)2J

or E=3.9×104J



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