Simple Harmonic Motion: Determining Half Potential and Half Kinetic Energy Displacement

Simple harmonic motion (SHM) is a common type of periodic motion where the restoring force is directly proportional to the displacement from the equilibrium position. A particle vibrating in SHM with an amplitude of 4cm requires careful analysis to understand the conditions under which the potential energy equals the kinetic energy. This article explores the mathematical processes involved, along with practical applications of SHM.

Understanding SHM and Energy in SHM

When a particle undergoes SHM, its total mechanical energy (E) is constant and can be expressed as the sum of kinetic energy (K) and potential energy (U). The potential energy (U) at a displacement (x) from the equilibrium position is given by:

U frac{1}{2} k x^2

where (k) is the spring constant.

Total Mechanical Energy in Terms of Amplitude

The total mechanical energy (E) can be expressed in terms of the amplitude (A). Therefore:

E frac{1}{2} k A^2

Given the amplitude (A 4 , text{cm} 0.04 , text{m}), the total energy is:

E frac{1}{2} k 0.04^2

Determining the Condition for Half Potential and Half Kinetic Energy

Now, we need to determine the displacement (x) from the equilibrium position at which the potential energy and kinetic energy are equal. Mathematically, this condition is expressed as:

U K

Since the total energy (E) is the sum of kinetic and potential energy, we can set up the equation:

E U K implies E 2U

This implies:

U frac{E}{2}

Substituting for (U), we get:

frac{1}{2} k x^2 frac{1}{2} E

Substituting (E), we obtain:

frac{1}{2} k x^2 frac{1}{2} left(frac{1}{2} k 0.04^2right)

Simplifying, we get:

k x^2 frac{1}{2} k 0.04^2

Dividing both sides by (k), assuming (k eq 0), we get:

x^2 frac{1}{2} 0.04^2

Calculating (x), we find:

x^2 frac{1}{2} 0.0016 0.0008

x sqrt{0.0008} approx 0.0283 , text{m} 2.83 , text{cm}

Thus, the displacement from the equilibrium position at which the particle has half potential and half kinetic energy is approximately (2.83 , text{cm}).

Alternative Method: A/Root 2 Approximation

This approximation can be remembered as A/sqrt{2}. For the given amplitude of (4 , text{cm}), the displacement is:

4/sqrt{2} 2sqrt{2} , text{cm}

Further Exploration of KE and PE in SHM

The kinetic energy (KE) and potential energy (PE) in SHM are given by:

KE frac{1}{2} m omega^2 (A^2 - x^2)

PE frac{1}{2} m omega^2 x^2

Where (m) is the mass of the particle, (omega) is the angular frequency, and (A) is the amplitude.

When PE equals KE, the equation becomes:

A^2 - x^2 x^2

This simplifies to:

2x^2 A^2

Thus, (x sqrt{frac{A^2}{2}} A/sqrt{2}). For an amplitude of (4 , text{cm}):

x frac{4}{sqrt{2}} 2sqrt{2} , text{cm}

Conclusion

This article has explored the critical aspect of determining the displacement in simple harmonic motion where the potential energy equals the kinetic energy. Through detailed mathematical analysis and practical applications, we have found the displacement to be approximately (2.83 , text{cm}) or (2sqrt{2} , text{cm}).