Question

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?

Answer 1

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