Can someone help me answer this SHM problem?

A block of mass m, rests on a frictionless horizontal surface. One end of a spring is connected to the block and the other end of the spring is connected to the wall. If the mass is displaced a distance x, it oscillates back and forth at a consistent frequency.

PE=1/2kx^2
If the block is displaced twice as much, what happens to the
A. energy of the system?
B. maximum velocity of the oscillating mass?
C. acceleration of the mass?

1 Answer
Jun 22, 2018

See below

Explanation:

The energy equation for this is:

  • E = T + U, where:

  • {(T = "kinetic energy" = 1/2 mv^2),(U = "potential (spring) energy" = 1/2 k x^2 ),(E = "total energy of the block" = 1/2 mv^2 + 1/2 k x^2):},

  • ...in each case, at time t.

E is constant through time, so the SHM involves energy switching periodically between T and U.

At maximum extension, x_("max"):

  • v = 0, qquad E = U_("max") = 1/2 kx_("max")^2

At zero extension, the equilibrium position:

  • x = 0, qquad E = T_("max") = 1/2 mv_("max")^2

Finally, using energy conservation again:

  • E = 1/2 kx_("max")^2 = 1/2 mv_("max")^2 = " const"

Now ANSWERING the questions:

A. energy of the system?

Because E = 1/2 kx_("max")^2, then E propto x_("max")^2.

Energy will quadruple.

B. maximum velocity of the oscillating mass?

1/2 kx_("max")^2 = 1/2 mv_("max")^2 implies abs (v_("max") )propto abs ( x_("max"))

Max velocity will double

C. acceleration of the mass?

For the spring:

  • F = - kx

From Newton's 2nd Law, for the mass:

  • bbF = mbba qquad = - kbbx

:. a = - k/m x, qquad implies abs(a_"max") propto abs(x_("max")

Acceleration through the equilibrium will still be zero, but the magnitude of acceleration at the extremities of the motion, when the mass is changing direction, will double