Question

cantly olne 763 n = 1, 2, 3, ... at frequency: mental frequency: 3) If 3. som of gor insedy STD inpbags unters preswe of 3.20tom. under lo 382 who pressure of 7.2 free of gar - Pui - Po V2 1/1) =P Vita 10m - 293. 15 dagre keln . Ta PT "/m? vor llam) 135m²) (380+ 273k) - 1.25m} an indecub of STP Units 227 kr A spring with stiffness constant 19.6 N/m is set up horizontally with the is set up horizontally with the 0.300 kg mass resting on a frictionless table. The mass is pulled so that the spring is stretched 0.100 m from the equilibrium point, and released from rest. (6 x 5 points) a) What is the amplitude of the horizontal oscillation A? b) What is the magnitude of the maximum velocity Vinu? c) What is the magnitude of the velocity V when the mass is 0.050 m from the equilibrium? d)) What is the total energy? e) What is the period of the oscillation? f) What is the frequency of the oscillation? d) & KA2 = x m=0.300 ks

Answer 1

Given, k =19.6 N/m

m = 0.3 kg

Amplitude, A = 0.1 m

displacement of the mass is given by,

x(t) = A*cos(wt)

w = sqrt(k/m)

w = sqrt(19.6/0.3)

w = 8.083 rad/s

a) A = 0.1 m

b) velocity. v = -Aw*sin(wt) = -w*sqrt(A² - x²)

v_max = Aw = 0.1*8.083 = 0.8083 m/s

c) v = -8.083*sqrt(0.1² - 0.05²)

v = -0.7 m/s

-ve simply represents the direction of the velocity

d) Total energy is the kinetic energy when speed is maximum,

E = 0.5mv_max² = 0.5*0.3*0.8083²

E = 0.098 J

e) Time period of oscillation, T = 2 $\pi$ /w

T = 2*3.14/8.083

T = 0.777 s

f) f = 1/T = 1/0.777

f = 1.287 Hz