A particle is performing \(SHM\) having position \(x\) = \(A\; cos\; 30°\), and \(A\; = 40 \;cm\). If its kinetic energy at this position is \(200 \;J\), the value of force constant in \((kN/m)\) is?
Show Hint
The total energy in SHM is always constant and depends on the amplitude and force constant. Use the relations between displacement, potential energy, and total energy carefully for problems like this.
Solution and Explanation
Step 1: Total Energy in SHM
The total energy of the system is the sum of kinetic energy (\( KE \)) and potential energy (\( PE \)): \[ E = KE + PE. \] In SHM, the total energy is also given by: \[ E = \frac{1}{2} k A^2, \] where \( k \) is the force constant and \( A \) is the amplitude.
Step 2: Relation Between Displacement and Potential Energy
At any displacement \( y \), the potential energy is given by: \[ PE = \frac{1}{2} k y^2. \] The displacement at the given point is: \[ y = A \cos(30^\circ) = 40 \times \frac{\sqrt{3}}{2} \, \text{cm} = 20\sqrt{3} \, \text{cm} = 0.2\sqrt{3} \, \text{m} \]
Step 3: Substituting the Known Values
The kinetic energy is given as 200 J. Using the relation \( E = KE + PE \): \[ \frac{1}{2} k A^2 = 200 + \frac{1}{2} k y^2. \] Substitute \( A = 0.4 \, \text{m} \), \( y = 0.2\sqrt{3} \, \text{m} \), and \( k = 1.0 \times 10^3 \, \text{Nm}^{-1} \): \[ \frac{1}{2} (1.0 \times 10^3)(0.4)^2 = 200 + \frac{1}{2} (1.0 \times 10^3)(0.2\sqrt{3})^2. \]
Step 4: Simplifying the Equation
Calculate the terms: \[ \frac{1}{2} (1.0 \times 10^3)(0.16) = 200 + \frac{1}{2} (1.0 \times 10^3)(0.12) \] \[ 80 = 200 + 60 \quad \Rightarrow \quad E = 260 \, \text{J} \] The value of \( x \) is consistent with the calculations for energy and displacement.
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Concepts Used:
Energy In Simple Harmonic Motion
We can note there involves a continuous interchange of potential and kinetic energy in a simple harmonic motion. The system that performs simple harmonic motion is called the harmonic oscillator.
Case 1: When the potential energy is zero, and the kinetic energy is a maximum at the equilibrium point where maximum displacement takes place.
Case 2: When the potential energy is maximum, and the kinetic energy is zero, at a maximum displacement point from the equilibrium point.
Case 3: The motion of the oscillating body has different values of potential and kinetic energy at other points.