Beat Frequency Problem

  • Thread starter nietzsche
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  • #1
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Homework Statement



Two strings with otherwise identical properties differ in tension by 0.25 N. If both strings are struck at the same time, what is the resulting beat frequency?

Homework Equations



frequency is proportional to the root of tension?

The Attempt at a Solution



I tried setting T2 = T1 - 0.25 N and substituting this into an equation with f1/f2 proportional to sqrt{T1/T2}.

But I can't seem to figure anything out from this.

Please help! Going a bit crazy.
 

Answers and Replies

  • #2
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I can't make anything of it either.

If I put [itex] f_1 = k \sqrt {T} [/itex] and

[tex] f_2 = k \sqrt {T + 0.25} [/tex]

I get

[tex] f_2 - f_1 = k \sqrt {T+0.25} - k \sqrt {T} [/tex]

wich I can't simplify further.

0.25 N is probably quite small compared to the tension in the string, so we can apply

[tex] \sqrt {T+0.25} \approx \sqrt {T} + 0.25 \frac {1} {2 \sqrt {T}} [/tex]

this results in

[tex] f_2 - f_1 \approx \frac {0.125 k} {\sqrt{T}} [/tex]

so the answer does depend on the tension, and the other properties of the string as well.
 
  • #3
186
0
I can't make anything of it either.

If I put [itex] f_1 = k \sqrt {T} [/itex] and

[tex] f_2 = k \sqrt {T + 0.25} [/tex]

I get

[tex] f_2 - f_1 = k \sqrt {T+0.25} - k \sqrt {T} [/tex]

wich I can't simplify further.

0.25 N is probably quite small compared to the tension in the string, so we can apply

[tex] \sqrt {T+0.25} \approx \sqrt {T} + 0.25 \frac {1} {2 \sqrt {T}} [/tex]

this results in

[tex] f_2 - f_1 \approx \frac {0.125 k} {\sqrt{T}} [/tex]

so the answer does depend on the tension, and the other properties of the string as well.

thank you so much willem. i don't think my prof worded the question properly, so i'm going to ask him tomorrow. thanks again!
 

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