# A problem about stationary waves ?

## Homework Statement

a violin string of length 40 cm and mass 1.2 gm has a frequency 500 hz when it produces fundamental tone where should you place your finger to increase frequency to 650 hz

## Homework Equations

frequency = n/2L . sqrt FT/ML

## The Attempt at a Solution

i just wanna make sure if the answer i got is right frequency 1 is 500 and frequency 2 is 650
i made v1/v2= n1/n2 x 2L2/2L1 x sqrt FT1/FT2 x sqrt ML2/ML1
so i crossed length 2 with length one due to length is the same and force of tension and mass per unit length also and remains the number of segments the first one n1 is 1 because its fundamental while the other one is unknown and he's asking for i think ?
so v1/v2 = n1/n2 500/650=n1/n2 since n1=1 so n2 = 10/13 = 1.3 m ? so i have to place my fingers 1.3 m away from the first node or 1.3 is the number of segments or what ? i just got stuck here s

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The question is about fundamental tones, so n = 1 in both case. When you place a finger on a string, you change its length. Everything else stays the same.

okay so i got L2
v1/v2=n1/n2 * 2L2/2L1 *√FT1/FT2 * √ML2/ML1
and i crossed mass per liter and forces of tension and number of segments so it's v1/v2= L2/L1
500/650=L2/40 L2=30.77cm but that's the length of the segment, still where should i place my finger on the string ??? I searched about the problem and found the final answer was L2 = 30.77 cm; "place finger 9.23 cm from scroll bridge" how did he knew i will place my fingers exactly 9.231 cm ? and what is meant by " from scroll bridge"

i know this feels stupid but if i placed my finger at the end of the string when it's 40 cm it will produce frequency of 500 hz ? while if i placed my finger at 9.23 cm away from the end it will prodce frequency of 650 ?

and if i did place my finger 9.23 away wouldn't the string make a second harmonic tone because it vibrates from the right and left of my finger ?

When you place a finger on a string, it won't be vibrating at the left and at the right. It will only vibrate where it is agitated with the bow. Another reason it won't vibrate on the other side is because most of it will be just pressed against the wood of the neck.

Second and higher order harmonics will obviously exist, but they will be multiples of the fundamental tone of the shortened string, not the original string.

ahhaaaaaa !! Got it! Thanks