1. Dec 8, 2003

### bard

At t=0, a 650-g(0.65kg) mass at rest on the end of a horizontal spring(k=184n/m) is struck by a hammer which gives it an initial speed of 2.26 m/s. Determine (a) the period and frequency of the motion (b)the amplitude (c)the maximum acceleration (d) the position as a function of time (e) the total energy, and (f)the kinetic energy when x=0.4A where A is the amplitude.

a) period=2pi*sqrt0.65/184=.37
Frequency=2.07
b)v=sqrtk/m(A^2-X^2)=0.13

C) A-max=184*0.13/0.65=36.8 m/s^2

d)x=0.13cos(phi)

e)no idea

f)no idea

2. Dec 8, 2003

### himanshu121

Revisit the formula for Total energy and K.E.

Total Energy remains Constant

3. Dec 8, 2003

### bard

so the total energy would be

1/2 KA^2

4. Dec 9, 2003

### himanshu121

yes and what about the K.E.

5. Dec 9, 2003

### HallsofIvy

Two other points:

I get that the frequency is 2.70 not 2.07. A typo?

The position, measured from the rest point, is .13 sin(16.8t).

There was no "phi" in the problem. What was that?

6. Dec 9, 2003

### bard

So the kinetic energy would be

1/2*184(0.052)^2=.25J