A 0.30 kg block oscillates back and forth along a straight line on a frictionless horizontal surface. Its displacement from the origin is given by x = (18 cm)cos[(11 rad/s)t + π/2 rad] (a) What is the oscillation frequency? (b) What is the maximum speed acquired by the block? (c) At what value of x does this occur? (d) What is the magnitude of the maximum acceleration of the block? (e) At what positive value of x does this occur? (f) What force, applied to the block by the spring, results in the given oscillation?
Given, x = (18 cm)cos[(11 rad/s)t + π/2 rad]
a.)
Standard equation is given by,
x = A*cos(w*t + π/2)
By comparing it with standard equation,
A = amplitude = 18 cm = 0.18 m
w = 11 rad/s
then, oscillation frequency = f = w/(2*pi)
f = 11/(2*pi)
f = 1.75 Hz
b.)
Since,
Maximum speed in oscillation is given by,
Vmax = A*w = 0.18*11
Vmax = 1.98 m/s
c.)
V = dx/dt = -A*w*sin(w*t + π/2)
V = -(18 cm)*(11 rad/s)*sin(11*t + π/2)
For maximum value,
sin(11*t + π/2) = 1
t = 0
d.)
Acceleration(a) = dV/dt = -A*w^2*cos(w*t + π/2)
So, maximum value will be,
a = A*w^2 = 0.18*11^2
a = 21.78 m/s^2
e.)
For maximum value of acceleration,
x = A
x = 18 cm
f.)
In oscillation:
w = sqrt(k/m)
k = m*w^2 = 0.30*11^2
k = 36.3 N/m
From hooke's law,
F = -k*x
F = -36.3*x
F = -36.3*(0.18*cos(11t + π/2))
F = -6.534*cos(11t + π/2)) N
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