When a player's finger presses a guitar string down onto a fret, the length of the vibrating portion of the string is shortened, thereby increasing the string's fundamental frequency. The string's tension and mass per unit length remain unchanged. If the unfingered length of the string is l=65cm, determine the positions x of the first six frets, if each fret raises the pitch of the fundamental by one musical note in comparison to the neighboring fret. On the equally tempered chromatic scale, the ratio of frequencies of neighboring notes is 21/12
x1=
x2=
x3=
x4=
x5=
x6=

Explanation:

For frequencies n generated in a string , the expression is as follows

n = 1 /2L√ ( T/m )

n is fundamental frequency , T is tension in string , m is mass per unit length and L is length of string.

If T and m are constant , then

n x L = constant , hence n is inversely proportional to L or length of string.

Frequencies increase by 21/12 = 1.75 , length must decrease by 1 / 1.75 times

Initial length of string is 65 cm .

x1 = 65 x 1 / 1.75 = 37.14 cm

x2 = 37.14 x 1/ 1.75 = 21.22 cm

x3 = 21.22 x 1 / 1.75 = 12.12 cm

x4= 12.12 x 1 / 1.75 = 6.92 cm

x5 = 6.92 x 1 / 1.75 = 3.95 cm

x6 = 3.95 x 1 / 1.75 = 2.25 cm

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.

the given statement is true.

sensory memory lasts for a fraction of seconds before the information is lost. it is the shortest -term memory and can be called as an element of memory. example: when we look at something and remember what it looks like, is just a second of observation. it degrades very quickly. it can also be considered as the process of perception rather than considering it as a memory.