Answer to Question #96448 in Classical Mechanics for Surej

Question

1.If K is a positive constant of the following expression represents simple harmonic
motion (x is the displacement of particle from mean position)
a. Acceleration = kx
b. acceleration = kx2
c. Acceleration =-kx
d. acceleration = –kx2
2.If k and a are the positive constants and x is the displacement from equilibrium
position of the following expression represents SHM.
a. Velocity = k(a2– x2)
b. Velocity = √k(x2– a2)
c. Velocity = √k(a2– x2)
d. Velocity = k(x2– a2)

Answer 1

Question 1.

Since for harmonic oscillator $U = {{\alpha {x^2}} \over 2}$ the equation of motion

$$m{{{d^2}\vec x} \over {d{t^2}}} = - \nabla U$$

has the form (there and furher ${\alpha \over m} = k$)

$${{{d^2}\vec x} \over {d{t^2}}} = - {\alpha \over m}x$$

Thus the right answer is C

Answer: C.

Question 2.

Let's write down the first integral of the equation

$${{{d^2}\vec x} \over {d{t^2}}} = - {\alpha \over m}x$$

It has the form $K + U = E$ where K is kinetic energy, U - potential energy and E is full energy. At the amplitude value of displacement $E = {U_0} = {{\alpha {A^2}} \over 2}$ and we have

$${{m{v^2}} \over 2} + {{\alpha {x^2}} \over 2} = {{\alpha {A^2}} \over 2}$$

Then

$$v = \pm \sqrt {{\alpha \over m}({A^2} - {x^2})}$$

The right answer is C.

Answer: C