A mass of 327 g is attached to a spring and set into simple harmonic motion with a period of 0.266 s. If the total energy of the oscillating system is 7.14 J, determine the following.
(a) maximum speed of the object
m/s
(b) force constant
N/m
(c) amplitude of the motion
m
NCS
Favorite Answer
(a) total mech energy = max KE = 7.14 J = ½ * 0.327kg * V²
so max V = 6.61 m/s
(b) from period T = 2π√(m/k) we rearrange to get
k = 4π²m / T² = 4π² * 0.327kg / (0.266s)² = 182 kg/s² = 182 N/m
(c) TME = max PE = 7.14 J = ½ * 182N/m * A²
so amplitude A = 0.280 m
Hope this helps!
oubaas
A mass of 327 g is attached to a spring and set into simple harmonic motion with a period of 0.266 s. If the total energy of the oscillating system is 7.14 J, determine the following:
(a) maximum speed of the object in m/s
7.14*2 = 0.266*V^2
V = √7.14*2/0.266 = 7.33 m/sec
(b) force constant k in N/m
0.266^2 = (2PI)^2*m/k
k = 6.2832^2*0.327/0.266^2 = 182.5 N/m
(c) amplitude of the motion
T = 2.0*x
x = T/2.0 = 0.266/2 = 0.133 m
Anonymous
uhh no i’m not doing all that. if it’s homework just write random numbers and make it look like you did work then you can get credit bc its a completion grade