What Frequency Does a String Vibrate at in Its Second Harmonic?

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A string fixed at both ends and measuring 0.29 m vibrates at its second harmonic with a wave speed of 115 m/s. The frequency of vibration can be calculated using the formula fn=n(v/2L), where n represents the harmonic number. For the second harmonic, n equals 2, leading to the equation 2f1=1(115 m/s / (2 * 0.29 m)). The confusion arises from correctly applying the formula and understanding the relationship between wavelength and string length. Visualizing the second harmonic can aid in grasping these concepts and solving the problem effectively.
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A string of length 0.29 m is fixed at both ends. The string is plucked and a standing wave is set up that is vibrating at its second harmonic. The traveling waves that make up the standing wave have a speed of 115 m/s. What is the frequency of vibration?



fn=n(v/2L)



I tried: 2f1=1(115m/s/2*0.29) then 2f1= 198.3 and I have both multiplied the answer by 2 and divided and both answers are wrong. The second harmonic is really throwing me for a loop. I am also confused about the n in the equation. Any help is greatly appreciated!
 
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If you are unsure about the formula, how about trying this approach: Try drawing what the second harmonic looks like, and then determine what the wavelength of the wave is in relation to the length of the string. See if you can work from there.
 
Thread 'Collision of a bullet on a rod-string system: query'

Thread 'Collision of a bullet on a rod-string system: query'

In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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