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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Simple Harmonic Motion

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Simple Harmonic Motion

Determine the elastic potential energy stored in each spring shown...

Determine the elastic potential energy stored in each spring shown in the figure when the block is in equilibrium. Additionally, determine the time period of vertical oscillation of the block.
Determine the elastic potential energy stored in each spring shown in the figure when the block is in equilibrium.

Options

  1. AEnergies: M2g2k1\dfrac{M^2 g^2}{k_1}, M2g2k2\dfrac{M^2 g^2}{k_2} and M2g2k3\dfrac{M^2 g^2}{k_3}; Time period: $2\pi \sqrt{M \left( \dfrac{1}{k_1} + \dfrac{1}{k_2} + \dfrac{1}{k_3} \right)}
  2. BEnergies: Mg22k1\dfrac{M g^2}{2 k_1}, Mg22k2\dfrac{M g^2}{2 k_2} and Mg22k3\dfrac{M g^2}{2 k_3}; Time period: 2πM(k1+k2+k3)2\pi \sqrt{M (k_1 + k_2 + k_3)}
  3. CEnergies: M2g22k1\dfrac{M^2 g^2}{2 k_1}, M2g22k2\dfrac{M^2 g^2}{2 k_2} and M2g22k3\dfrac{M^2 g^2}{2 k_3}; Time period: 2πMk1+k2+k32\pi \sqrt{\dfrac{M}{k_1 + k_2 + k_3}}
  4. DEnergies: M2g22k1\dfrac{M^2 g^2}{2 k_1}, M2g22k2\dfrac{M^2 g^2}{2 k_2} and M2g22k3\dfrac{M^2 g^2}{2 k_3}; Time period: $2\pi \sqrt{M \left( \dfrac{1}{k_1} + \dfrac{1}{k_2} + \dfrac{1}{k_3} \right)}

Correct answer

D. Energies: M2g22k1\dfrac{M^2 g^2}{2 k_1}, M2g22k2\dfrac{M^2 g^2}{2 k_2} and M2g22k3\dfrac{M^2 g^2}{2 k_3}; Time period: $2\pi \sqrt{M \left( \dfrac{1}{k_1} + \dfrac{1}{k_2} + \dfrac{1}{k_3} \right)}

Step-by-step solution

In equilibrium, the tension in each spring is equal to the weight of the block, MgMg.

The extension in each spring is given by:
x1=Mgk1x_1 = \dfrac{Mg}{k_1}, x2=Mgk2x_2 = \dfrac{Mg}{k_2}, x3=Mgk3x_3 = \dfrac{Mg}{k_3}

The elastic potential energy stored in each spring is U=12kx2U = \dfrac{1}{2} k x^2.
U1=12k1(Mgk1)2=M2g22k1U_1 = \dfrac{1}{2} k_1 \left(\dfrac{Mg}{k_1}\right)^2 = \dfrac{M^2 g^2}{2 k_1}
U2=12k2(Mgk2)2=M2g22k2U_2 = \dfrac{1}{2} k_2 \left(\dfrac{Mg}{k_2}\right)^2 = \dfrac{M^2 g^2}{2 k_2}
U3=12k3(Mgk3)2=M2g22k3U_3 = \dfrac{1}{2} k_3 \left(\dfrac{Mg}{k_3}\right)^2 = \dfrac{M^2 g^2}{2 k_3}

Since the springs are connected in series, the equivalent spring constant keqk_{eq} is given by:
1keq=1k1+1k2+1k3\dfrac{1}{k_{eq}} = \dfrac{1}{k_1} + \dfrac{1}{k_2} + \dfrac{1}{k_3}

The time period of vertical oscillation is:
$T = 2\pi \sqrt{\dfrac{M}{k_{eq}}} = 2\pi \sqrt{M \left( \dfrac{1}{k_1} + \dfrac{1}{k_2} + \dfrac{1}{k_3} \right)}
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