An undamped 1.24 kg horizontal spring oscillator has a spring constant of 23.2 N/m. While oscillating, it is found to have a speed of 2.84 m/s as it passes through its equilibrium position. What is its amplitude A of oscillation? A Incorrect What is the oscillator’s total mechanical energy E
tot
?
as it passes through a position that is 0.676 of the amplitude away from the equilibrium position? E
The correct answer and explanation is:
To solve the problem, let’s first find the amplitude AA and then calculate the total mechanical energy at a specific position during the oscillation.
Part 1: Finding the Amplitude AA
The mechanical energy in a spring oscillator is given by the formula: Etot=12kA2E_{\text{tot}} = \frac{1}{2} k A^2
where:
- kk is the spring constant (23.2 N/m),
- AA is the amplitude of the oscillation.
At the equilibrium position (where x=0x = 0), the speed vv is at its maximum, and all of the mechanical energy is kinetic. The kinetic energy is: Ekin=12mv2E_{\text{kin}} = \frac{1}{2} m v^2
where:
- mm is the mass of the oscillator (1.24 kg),
- vv is the speed at the equilibrium position (2.84 m/s).
Since the total mechanical energy in the system is conserved, we know that the total energy is equal to the maximum kinetic energy: Etot=Ekin=12mv2E_{\text{tot}} = E_{\text{kin}} = \frac{1}{2} m v^2
Substitute the given values: Etot=12(1.24 kg)(2.84 m/s)2E_{\text{tot}} = \frac{1}{2} (1.24 \, \text{kg}) (2.84 \, \text{m/s})^2
Now we can calculate EtotE_{\text{tot}}: Etot=12(1.24)(8.0656)=5.0 JE_{\text{tot}} = \frac{1}{2} (1.24) (8.0656) = 5.0 \, \text{J}
Now, we use the total mechanical energy formula Etot=12kA2E_{\text{tot}} = \frac{1}{2} k A^2 to solve for AA: 5.0=12(23.2)A25.0 = \frac{1}{2} (23.2) A^2 A2=10.023.2A^2 = \frac{10.0}{23.2} A2=0.431⇒A=0.431≈0.656 mA^2 = 0.431 \quad \Rightarrow \quad A = \sqrt{0.431} \approx 0.656 \, \text{m}
So, the amplitude of oscillation AA is approximately 0.656 m.
Part 2: Total Mechanical Energy at 0.676 of the Amplitude Away from the Equilibrium Position
The position where the oscillator is 0.676 of the amplitude away from the equilibrium position corresponds to x=0.676Ax = 0.676A. The potential energy at this position is: Epot=12kx2E_{\text{pot}} = \frac{1}{2} k x^2
Substituting x=0.676Ax = 0.676A: Epot=12k(0.676A)2E_{\text{pot}} = \frac{1}{2} k (0.676A)^2
The total mechanical energy EtotE_{\text{tot}} is conserved, so it remains 5.0 J, and the sum of potential and kinetic energy at any position is equal to this total energy.
Thus, the kinetic energy at this position is: Ekin=Etot−EpotE_{\text{kin}} = E_{\text{tot}} – E_{\text{pot}}
Since Etot=5.0 JE_{\text{tot}} = 5.0 \, \text{J}, let’s calculate the potential energy when x=0.676Ax = 0.676A using the formula: Epot=12(23.2)(0.676×0.656)2E_{\text{pot}} = \frac{1}{2} (23.2) (0.676 \times 0.656)^2
We can now compute the energy values and final results.
Final Explanation
The amplitude was determined by using the maximum kinetic energy at equilibrium and equating it to the total mechanical energy in the system. By relating the total mechanical energy to the spring constant and amplitude, we derived the amplitude value. Then, by using the same energy concepts, we calculated the total mechanical energy when the oscillator is a certain distance away from the equilibrium. This method demonstrates the conservation of mechanical energy in simple harmonic motion.