1.2/3 points I An object of mass m is suspended from a vertical spring of force

Question

1.2/3 points I An object of mass m is suspended from a vertical spring of force constant 1746 N/m. When the object is pulled down 2.29 cm from equilibrium and released from rest, the object oscillates at 5.30 Hz. Notes Ask Your Te (a) Find m. m = 1 .58 v kg (b) Find the amount the spring is stretched from its unstressed length when the object is in equilibrium. ( = 8.87 (c) Write expressions for the displacement x, the velocity Vxi and the acceleration a, as functions of time t. n--2,290cosl(36,7.xm vx(t) = cm/s ax(t)- m/s2 You can find the mass of the object from the force constant and oscillation frequency. You can apply a condition for translational equilibrium to the object when it is at its equilibrium position to determine the amount the spring has stretched from its natural length. Finally, use the initial conditions to determine the amplitude and phase constants, and then differentiate this expression to obtain expressions for the x components of the velocity and acceleration. 14.P.062 If the period of a 74.6-cm-long simple pendulum is 1.51 s, what is the value of g at the location of the 12.91 m/s2

Explanation / Answer

mass = m

k = 1746 N/m

f = 5.3 Hz

A = 2.29 cm

a. now w = 2*pi*f = sqroot(k/m)

4*pi^2*f^2 = 1746/m

m = 1.5744 kg

b. xo = mg/k = 0.884 cm

c. x(t) = -Acos(2*pi*f*t)

x(t) = -2.29cos(33.3t) cm

v(t) = dx/dt = 76.257sin(33.3t) cm/s

a(t) = 2539.3581cos(33.3t) cm/s/s

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