Question #46523

1 Answer
Mar 27, 2017

#20cm#

Explanation:

Assuming that we are referring to the fundamental vibrational mode of the stretched string.

We know that in such a case the wavelength of the note is twice the length of the string.
#L=lambda/2# ......(1)
It can be also be shown that
Wave velocity #v=sqrt(T/rho)#
where #T# is tension in the string and #rho# is mass per unit length of wire.

Using the wave equation and #L# as length of string, we get
Frequency #f_1=v/lambda=sqrt(T/rho)/(2L)# .....(2)

Case 1. Given #256=sqrt(T/rho)/(2xx80)#
#=>sqrt(T/rho)=256(2xx80)# .....(3)

case 2. To find #1024=sqrt(T/rho)/(2xxL)#
Since it is similar wire under similar tension, using (3) we get
#1024=(256(2xx80))/(2xxL)#
#=>L=(256(2xx80))/(2xx1024)#
#=>L=20cm#