- #1
Karol
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Homework Statement
A wire of length 1[m] vibrates with the base frequency which is 200[Hz]. the specific mass is 8[gr/cm3].
The maximal acceleration at the middle is 80,000[cm/sec2]. what is the amplitude
Homework Equations
Newton's second law: [itex]F=ma[/itex]
The cosine sentence: [itex]A^2=B^2+C^2-2BC\cdot \cos \alpha[/itex]
The speed of transverse waves in a wire: [itex]u=\sqrt{\frac{P}{\rho}}[/itex]
P is the stress and [itex]\rho[/itex] is the specific mass.
The speed as a function of wavelength and frequency: [itex]u=\lambda f[/itex]
The Attempt at a Solution
The tension is the stress times the area. and the force i take from Newton's law:
##F=100[cm]\cdot 8[gr/cm^3]\cdot a \cdot 80,000=64E6\cdot A[dyn]##
The velocity:
##u=\lambda f\rightarrow 200=\frac{u}{2\cdot 100}\rightarrow u=40,000[cm/s]##
The stress:
##u=\sqrt{\frac{P}{\rho}}\rightarrow 40,000=\sqrt{\frac{P}{8}}\rightarrow P=200E6[dyn/cm^2]##
I drew the forces in the stretched wire, i find the resultant force:
##F^2=2T^2-2T^2\cos 2\alpha##
##(64E6)^2\cdot A^2=(200E6)^2A^2(1-\cos 2\alpha)##
##\rightarrow \alpha=9.21^0 \rightarrow h=8.1[cm]##
It's a little high, no? is my solution correct?