Fundamental frequency of a stretched string

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Homework Help Overview

The problem involves determining the new fundamental frequency of a stretched string when its length is doubled and the tension is increased significantly. The original frequency is given as 200 Hz.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of the formula for fundamental frequency and question its correctness, particularly regarding the variables used and their implications for the calculation.

Discussion Status

There is an ongoing examination of the formula used for calculating frequency, with some participants suggesting that the original poster's answer may be correct despite concerns about the provided options. Multiple interpretations of the problem and its parameters are being explored.

Contextual Notes

Participants note potential discrepancies in the problem's setup and the provided answers, indicating a lack of clarity in the question itself. There is also discussion about the notation used for linear density versus mass.

lykan_004
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Homework Statement



The fundamental frequency of a stretched string is 200Hz. when the length of the string is doubled and Tension of the string made 100times the initial Tension, what is the new fundamental frequency of the string.

(1) 50 Hz (2) 100 Hz (3) 200Hz (4) 400 Hz (5) 800 Hz

Homework Equations



F = 1/2L x sqrt(T/ m)

The Attempt at a Solution



The answer I get is 1000Hz.
 
Last edited:
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correct attempt
 
does that mean the question is wrong? coz there are similar questions, and i don't get the correct answer for any of them.
 
lykan_004 said:

Homework Equations



F = 1/2L x sqrt(T/ m)

I think you'll want to verify this formula. It seems to be missing a factor of L inside the square root. You can verify by checking that the units don't work out to Hz in its present form.
 
gneill said:
I think you'll want to verify this formula. It seems to be missing a factor of L inside the square root. You can verify by checking that the units don't work out to Hz in its present form.

m - linear density here not mass..that makes the equation dimensionally correct.
 
lykan_004 said:
m - linear density here not mass..that makes the equation dimensionally correct.

Ah. Perhaps then \rho would have been a better choice of variable name :smile:

So it would appear that your answer is correct; The frequency should change by a factor of \sqrt{100}/2 = 5. It sometimes happens that the provided answers are not correct.
 
u r rite .. i wonder the same thing... :) but i have no idea why they use m.
 

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