Question

Two stretched strings are placed next to one another, one with a length of 70.0 cm and the other with a length of 84.0 cm. The two strings have the same mass per unit length. You pluck the shorter one so that it vibrates at its fundamental frequency. You then adjust the tension in the longer string until it resonates with the first one. To resonate, the two strings must have the same fundamental frequency, so that vibrations on one string can cause the second string to vibrate. (a) What is the ratio of the speed of waves on the shorter string to the speed of waves on the longer string? (b) What is the ratio of the tension in the shorter string to the tension in the longer string?

Answer 1

`(v_s)/(v_l)=(5)/(6)=0.8333...`

`(T_s)/(T_l)=(25)/(36)=0.69444...`

Step-by-step explanation:

The frequency of the nth harmonic on a string is given by the formula:

`f_n=(nv)/(2L)`

which can be written as:

`v=(2Lf_n)/(n)`

Since the fundamental frequency means n=1, the ratio of the speed of waves on the shorter string (
`v_s`) to the speed of waves on the longer string (
`v_l`) will be:

`(v_s)/(v_l)=(2L_sf_(1s))/(2L_lf_(1l))=(L_s)/(L_l)`

Since the frequencies are the same, and for our values is:

`(v_s)/(v_l)=(L_s)/(L_l)=(70cm)/(84cm)=(5)/(6)=0.8333...`

The speed of the waves relates to the tension and mass per unit length by the formula:

`v=\sqrt{(T)/(\lambda)}`

which can be written as:

`T=v^2 \lambda`

The ratio of the tension in the shorter string to the tension in the longer string will be:

`(T_s)/(T_l)=(v_s^2 \lambda_s)/(v_l^2 \lambda_l)=(v_s^2)/(v_l^2)=((v_s)/(v_l))^2`

Since the mass per unit length are the same, and for our values is:

`(T_s)/(T_l)=((v_s)/(v_l))^2=((5)/(6))^2=(25)/(36)=0.69444...`