Amplitude Question Just Need A Formula

[MJC8719]

At t = 0, a 730 g mass at rest on the end of a horizontal spring (k = 128 N/m) is struck by a hammer, which gives it an initial speed of 2.74 m/s.
(a) Determine the period of the motion.
s
Determine the frequency of the motion.
Hz
(b) Determine the amplitude.
m

I have found parts A with no problem...but i was hoping someone could post the formula that I can use to find the amplitude as I am having some real trouble figuing it out.

 

[G01]

Please show some work. You won't get help until you do, according to the forum guidelines. (I don't mean to be rude, but it's the only way we can be sure the person asking for help doesn't just want someone else to do their homework.)

 

[m2003]

The formula for amplitude in simple harmonic motion is given by A = x_max - x_0, where x_max is the maximum displacement from equilibrium and x_0 is the equilibrium position. In this case, the equilibrium position is where the mass is at rest, so x_0 = 0. The maximum displacement can be found by using the equation for the position of a mass on a spring, x = A*cos(ωt), where A is the amplitude and ω is the angular frequency. We can rearrange this equation to solve for A, giving us A = x_max = x/(cos(ωt)). Plugging in the values given in the problem, we get A = (0.730 kg * 2.74 m/s)/(cos(sqrt(k/m)*t)). The square root of k/m is the angular frequency, ω, which can be found using the formula ω = sqrt(k/m). Plugging in the values for k and m, we get ω = sqrt(128 N/m / 0.730 kg) = 12.17 rad/s. Finally, plugging in this value for ω and the given time, t = 0, we get A = (0.730 kg * 2.74 m/s)/(cos(12.17 rad/s * 0 s)) = 0.730 kg * 2.74 m/s = 2.00 m. Therefore, the amplitude of the motion is 2.00 m.