A mass is hanging on a vertical spring, and oscillates in SHM with an amplitude

Question

A mass is hanging on a vertical spring, and oscillates in SHM with an amplitude of 10.0 cm, and period 0.500 s. The graph shows its motion as a function of time. At t= 0, the mass is found at x= -7.50 cm below the equilibrium position.

(a) Assuming that x(t) = A cos (?t + ?), find the value of the initial phase angle.

(b) Calculate the velocity( magnitude and direction ) at t=0s

(c) Calculate the acceleration in m/s^2 at t=0.

(d) Is the speed incresing or decreasing at t=0? Explain briefly.

(e) Calculate the values of t corresponding to the points P ( mass at equilibrium ) and Q ( mass at maximum amplitude ) on the graph. Show your work.

A mass is hanging on a vertical spring, and oscillates in SHM with an amplitude of 10.0 cm, and period 0.500 s. The graph shows its motion as a function of time. At t= 0, the mass is found at x= -7.50 cm below the equilibrium position. Assuming that x(t) = A cos (?t + ?), find the value of the initial phase angle. Calculate the velocity( magnitude and direction ) at t=0s Calculate the acceleration in m/s^2 at t=0. Is the speed increasing or decreasing at t=0? Explain briefly. Calculate the values of t corresponding to the points P ( mass at equilibrium ) and Q ( mass at maximum amplitude ) on the graph. Show your work.

Explanation / Answer

x(t) = A cos (?t + ?)

given A = 10 cm
x(0) = -7.5

so.. x(0) = A * cos ( 0 + ?)
so.. -7.5 = 10 cos ( ?)
so.. ? = 180 + cos inverse(0.75) = 221.4096 degree = 3.8643 radians


period = 0.5 sec
so.. ? = 2*pi / time = 2*pi / 0.5 = 4*pi

b) so.. x(t) = A cos (?t + ?)
so. velcotity = dx / dt
so.. v(t) = -A*? sin (?t + ?)

so.. at t = 0 ..
velcity = v(0) = -A*w*sin (3.8643 ) = -0.1 *4pi *sin (3.8643 ) = 0.831187 m/sec

c) acceletaion = a(t) = dv/dt = -A*?^2 cos (?t + ?)
at t= 0
a(0) = -A*w^2*cos (3.8643 ) = -0.1 *(4pi)^2 *cos (3.8643 ) = 11.8435 m/sec2

d) at = 0 .... acceleration is positive and velocity is also positive.. so the speed is increaseing at t = 0

e)

so.. x(t) = 0.1 cos (4*pi*t + 3.8643)
0 = 0.1 cos (4*pi*t + 3.8643)
so. cos (4*pi*t + 3.8643) = 0
so.. 4*pi*t + 3.8643 = (n + 0.5 )*pi
so.. 4*pi*t = n*pi + 0.5*pi
so... time t = (pi* n - 2.2935) / ( 4pi)

for n = 0 ... t < 0 .. not possible..

so. n =1 .. t = 0.067489 secs <--------------- point P
for n = 2.. t = 0.317489 secs < --------------- point Q

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