Learning object for SHM PHYS 101 UBC

1. Simple Harmonic Motion of a Violin String
Open A on the Violin has a frequency of 440 Hz. If the A string is plucked so
its initial location of oscillation is π/2, and it oscillates with an amplitude of
1.0mm, what would the velocity of the string be after 2.0 seconds?
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We know that the equation for Velocity is: V(t)=(ωA)sin(ωt+Ø)
1) Write out your knowns
A= 1mm, or 0.001 m
t= 2.0 s
Ø=π/2
f= 440 Hz
T=?
ω=?
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2. 2) Find your unknowns
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The period, T, can be found with the equation T=1/f, and because we know
that f is 440Hz, it’s easy to solve for the period!
T=1/440
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ω is equal to 2π/T, so just sub in our value for T.
ω= 2π/(1/440)
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3) Sub into our equation for velocity
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V(t)=(ωA)sin(ωt+Ø)
V(2.0s)=
[(2π/(1/440)rad/second)(0.001 m)] x sin[(2π/(1/440)rad/second)(2.0s)+(π/2)]
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This gives us V(2.0)=2.76460 m/s OR 2760 mm/s…..That’s pretty fast!
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4) Rationalize your answer
Look at a simple graph of Sin, try to think of all of the variables we are using
in terms of this graph.
Ø describes different possible initial locations of the oscillation (in our
example this is π/2) and A is the amplitude of the curve (1 mm). We Pulled the
string 1 mm, to it’s position of π/2 to start the oscillation.
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