Two simple pendulums of length(l) are suspended from a ceiling(their strings are parallel). Their bobs, each of mass(m), are connected with a rubber string of force constant k. Assume the bobs are pulled away from each other, only by a little, so that the rubber band is stretched. When they are released simultaneously, what would be the time-period of each bob? (Neglect gravity between bobs and air resistance, and assume that the rubber string obeys Hooke's law, activity is performed in place with acceleration due to gravity, g.)




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   Please see diagram.

The pendulums are each displaced by an amount x.  The string is stretched by 2x totally.  The restoration force in the spring = F2 = k * 2x = 2k x.  (Hooke's law).
  Bob's displacement X in the tangential direction to the string = X
   x = displacement in the horizontal direction = X Cos Ф
   x  = L sin Ф

   mg CosФ = Tension T in the string
   F1 = component of weight  along the tangential direction
     F1 = mg SinФ = m g x / L

   Net force on the bob = F  = m a = - m * d²X / dt²
           (-ve sign is because the force is in the decreasing x direction)

   X = x / cos Ф = L tan Ф  ≈ L Sin Ф ≈ L Ф,   approximation for small Ф.

   Net force on pendulum in the tangential direction = \vec{F} = \vec{F1} + \vec{F2}

    F1 and F2 are at an angle Ф, and their vector sum is to be calculated for the resultant force and acceleration.  But since Ф is small, we approximate the resultant force to be along horizontal direction and F1 and F2 along this direction.  So we take the linear sum for simplicity.

    a =  - d² X / dt²  ≈  - d² x / dt²
      =>   - m d² x / dt²  =  2k x + m g x/L
  Resultant force on the bob =  mg x/L + 2 k x = F = m a = - m d^2 x/dt^2

             d² x/dt²  = - (g/L + 2k/m) x 

The equation of motion above indicates that for small amplitudes x and displacements, the  pendulum oscillates in a SHM and  the corresponding

      Angular velocity = ω = √(g/L + 2k/m)= \sqrt{\frac{g}{L}+\frac{2k}{m}}

      Time period = 2\pi\ \sqrt{\frac{mL}{mg+2kL}}
If displacements are large then, the resultant force F will be:
m\frac{d^2x}{dt^2}=\sqrt{F_1^2+F_2^2+2F_1\ F_2\ Cos\phi}\\\\=\sqrt{\frac{m^2g^2}{L^2}+4k^2+\frac{4kmk}{L}Cos\phi}\ *\ x
1 5 1
T = pi sqroot [ mL / (mg+2kL) ] + pi sqroot (L/g)
left Half swing of the left side pendulum will be as described above in the answer. The right half swing of the right side pendulum will be as the usual gravity pendulum.
the altitude h that left side pendulum goes up on the right side is found by energy conservation: 2 m g x tanФ + 1/2 k (2x)^2 = 2 m g h ; => hence h = x tanФ + k x^2 /mg.. or, h = L(1-cosФ)+k x^2 /mg
Let g’ = g + 2kL/m
T = π [ √( L/g’) + √(L/g) ]
Angular velocity = ω = 2 / [ √( L/g’) + √(L/g) ]
these are the correct answers.