Answer to Question #83101 in Quantum Mechanics for Naima

Question

1. The amplitude and frequency of an object executing simple harmonic motion are 0.01 m and 12 Hz respectively. What is the velocity of object at displacement 0.005 m? What is the maximum velocity of the object?
2. If the length of second’s pendulum is increased by25.6%, find the period of it.
3. The equation of motion of particle executing SHM is

Answer 1

1. The equation of simple harmonic motion can be written in the form x(t) = A sin (ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase. The angular frequency is related to the ordinary frequency f by the formula ω = 2πf. Now, x(t) is the displacement of object, while its velocity is given by the derivative v(t)=x'(t)=Aωcos(ωt+φ)=±Aω√(1-〖sin〗^2 (ωt+φ) )=±ω√(A^2-x^2 (t) ). The sign here depends on the phase of the motion. The absolute velocity v at a given displacement x is thus given by v=ω√(A^2-x^2 )=2πf√(A^2-x^2 ). Substituting here A = 0.01 m, x = 0.005 m, and f = 12 Hz = 12 s–1, we obtain, approximately, v = 0.653 m/s. The maximum velocity vm is reached at x = 0 and is given by vm = 2πfA = 0.754 m/s. Answer: v = 0.653 m/s, vm = 2πfA = 0.754 m/s.

2 The period of a simple pendulum is given by T=2π√(L⁄g), where L is its length, and g is the acceleration of gravity. If the length is increased by 25.6%, it is thus increased 1.256 times, hence, its period is increased √1.256≈1.12 times, and thus constitutes approximately 1.12 seconds. Answer: approximately 1.12 seconds.

3. The equation of motion of particle executing simple harmonic motion (SHM) is x(t) = A sin (ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase.