what is the displacement of an object in shm when kinetic and potential energies are equal?
Answers
Answer:x= ±12–√A.
Explanation: Consider the oscillator to be a mass, m, on a spring of constant, k. Use conservation of mechanical energy, E=KE+PE, where, KE=12mv2 and PE=12kx2. Since E is constant, its value is just the maximum value of the potential energy, which occurs when the velocity is zero, and the displacement is x=±A. Therefore, E=12kA2, where A is the amplitude. Given that KE=PE, one has
E=2PE. Consequently, E=kx2. Thus,
12kA2=kx2.
Therefore,
x2=12A2.
Taking the square root of both sides gives,
Consider the oscillator to be a mass, m, on a spring of constant, k. Use conservation of mechanical energy, E=KE+PE, where, KE=12mv2 and PE=12kx2. Since E is constant, its value is just the maximum value of the potential energy, which occurs when the velocity is zero, and the displacement is x=±A. Therefore, E=12kA2, where A is the amplitude. Given that KE=PE, one has
E=2PE. Consequently, E=kx2. Thus,
12kA2=kx2.
Therefore,
x2=12A2.
Taking the square root of both sides gives,
x=±12–√A.