Question
A small mass attached to a spring oscillates with simple harmonic motion with amplitude of 70mm,taking 13 seconds to make 40 complete oscillations.Calculate the following showing steps;
a.Its angular ferquency
b.Its maximum speed
c.Its maximum acceleration
d.Its speed when displacement from equilibrium is 10mm
e.Its speed when displacement from equilibrium is 20mm
a.Its angular ferquency
b.Its maximum speed
c.Its maximum acceleration
d.Its speed when displacement from equilibrium is 10mm
e.Its speed when displacement from equilibrium is 20mm
Answers
ΔT = t/N = 13/40 = 0.325 s.
a.
ω = 2•π/T = 2•π/0.325 = 19.33 rad/s.
b.
v(max) = A•ω= 70•10^-3•19.33 = 1.353 m/s
c.
a(max) =A•ω^2 = 26.16 rad/s^2.
d.
x= A•cos(ω•t+α) = 10 mm
cos(ω•t+α) = x/A= 10/70 =1/7.
sin(ω•t+α) = sqrt(1 – (cos(ω•t+α))^2) = 0.9897,
v = - A• ω• sin(ω•t+α) = 70•10^-3•19.33•0.9897 = -1.34 m/s.
e. similar to "d".
a.
ω = 2•π/T = 2•π/0.325 = 19.33 rad/s.
b.
v(max) = A•ω= 70•10^-3•19.33 = 1.353 m/s
c.
a(max) =A•ω^2 = 26.16 rad/s^2.
d.
x= A•cos(ω•t+α) = 10 mm
cos(ω•t+α) = x/A= 10/70 =1/7.
sin(ω•t+α) = sqrt(1 – (cos(ω•t+α))^2) = 0.9897,
v = - A• ω• sin(ω•t+α) = 70•10^-3•19.33•0.9897 = -1.34 m/s.
e. similar to "d".
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