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Unformatted text preview: MC g = MCω 2 x where x is the displacement from equilibrium, as in part (a). µs g
2
f max =
€
4 π 2 x max
i.e. µsg 1 / 2 0.5 × 9.8 1 / 2
1
∴ f max = 2 = 1.369s = 1.4 Hz to 2SF = 2 4 π 0.0663 4 π x max € € Higher Physics 1A € € € € PHYS1131 2010T1, Solutions Q5 23 Marks (a) The wave speed of a mechanical wave is related to the elastic properties and the inertial properties of the medium through which it is travelling, and is given by: Elastic Property
v= Inertial Property For a stretched string, of tension T and mass per unit length µ, these combine to give v = T / µ . (b) (i) We have a wave of form y = A sin( kx − ωt ) Amplitude A=0.200mm = 0.000200m €
Frequency f=500 Hz with ω=2πf = 1000π rad/s = 3,142 rad/s Wavespeed v=196 m/s=ω/k so that k=ω/v=2πf/v=1000π/196=16.03 m‐1 Thus, the wave equation becomes: y = 0.00200sin(16.0 x − 3140 t ) m to 3SF. (ii)We have v = T / µ so that T = µv 2 = 4.10 × 10−3 × 196 2 N = 157.5 N = 158 N to 3SF. (c) € Let the Tension be T at a distance x from the end, as in the diagram. Then T = µxg = Weight of string below x . Thus, wave speed at x is given by: € v = T /µ =
dx
= gx .
dt
x dx
So ∫
=
0
gx µxg
= xg .
µ ∴ € € € ∫ τ
0 dt i.e. t = 2 x / g is the time to traverse a distance x along the string, since at t=0 we have x=0. So when x=L, τ = 2 L / g is the time for the pulse to traverse the length of the string. (d) € Suppose we add a mass M to the bottom of the string, as in the diagram above...
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This note was uploaded on 12/02/2012 for the course PHYS 1131 taught by Professor Angstman during the Three '12 term at University of New South Wales.
 Three '12
 Angstman
 Physics, Work

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