A guitar string vibrates at 400 Hz and has a length of 0.8 m. Find. A guitar string vibrates at 400 Hz and has a length of 0.8 m. Find the new frequency if

a. The string is lengthened to 0.9 m.

b. The string is replaced with another string, the density of which is greater by a factor of two.

c. The tension in the string is decreased by a factor of 0.4.

d. The string is replaced with another string, the diameter of which is twice as large.

Answer:

(a) 355.56 Hz

(b) 282.84 Hz

(c) 309.84 Hz

(d) 200 Hz

Step-by-step explanation

n₁ = primary frequency = 400 Hz

L₁ = primary length = 0.8 m

(a)

New length = L₂ = 0.9 m

Frequency of the length is string isproportional to the length of the string. i.e ?n∝1/L??

Therefore,

?n2/n1???=L1/L2?

??or,  

n2?=0.8/0.9?×400?

= 355.56 Hz

(b) ?n∝1/ρ??? , ?ρ? is the density

Here, ?ρ2?=2 ρ1??

Therefore,

??n2/n1??=underoot ρ1/ρ2????

or,  

n2?=1/2??×400?

= 282.84 Hz

(c)

?n∝wnderootT?? , T is the tension

T₂ = T₁ ??- 0.4 T₁

= 0.6 T₁

Now, ??n2/n1??=underootT2/T1????

or,  

n2?=underoot0.6?×400?

= 309.84 Hz

(d) ?n∝1/d?? , where d is the diameter

Here, d₂ = 2 d₁

Now, ?n2/n1???=?d1/d2

??or, 

n2?=1/2?×400?

= 200 Hz