A simple harmonic oscillator has a total energy of 50 J and an amplitude of 10 cm. What is the spring constant?
Practice Questions
1 question
Q1
A simple harmonic oscillator has a total energy of 50 J and an amplitude of 10 cm. What is the spring constant?
200 N/m
500 N/m
1000 N/m
2000 N/m
Total energy E = (1/2)kA^2. 50 = (1/2)k(0.1^2) => k = 500 N/m.
Questions & Step-by-step Solutions
1 item
Q
Q: A simple harmonic oscillator has a total energy of 50 J and an amplitude of 10 cm. What is the spring constant?
Solution: Total energy E = (1/2)kA^2. 50 = (1/2)k(0.1^2) => k = 500 N/m.
Steps: 8
Step 1: Understand that the total energy (E) of a simple harmonic oscillator is given by the formula E = (1/2)kA^2, where k is the spring constant and A is the amplitude.
Step 2: Identify the values given in the problem. The total energy E is 50 J and the amplitude A is 10 cm. Convert the amplitude from cm to meters: 10 cm = 0.1 m.
Step 3: Substitute the known values into the formula: 50 = (1/2)k(0.1^2).
Step 4: Calculate (0.1^2), which is 0.01.
Step 5: Rewrite the equation with the calculated value: 50 = (1/2)k(0.01).
Step 6: Multiply both sides of the equation by 2 to eliminate the fraction: 100 = k(0.01).
Step 7: Divide both sides by 0.01 to solve for k: k = 100 / 0.01.
