Question

For a string stretched between two supports, two successive standing-wave frequencies are 535 Hz and 642 Hz. There are other standing-wave frequencies lower than 535 Hz and higher than 642 Hz.

Answer 1

Answer:

The length of the string is 1.73 m.

Explanation:

Given that,

Successive frequency f= 535 Hz

Successive frequency f' =642 Hz

Suppose, if the speed of transverse waves on the string is 372 m/s , what is the length of the string? Assume that the mass of the wire is small enough for its effect on the tension in the wire to be neglected.

The nth harmonic is

$f_{n}=\dfrac{nv}{2L}$....(I)

The (n+1)th harmonic is

$f_{n+1}=\dfrac{(n+1)v}{2L}$....(II)

We need to calculate the length of the string

Subtract equation (I)  from equation(I)

$f_{n+1}-f_{n}=\dfrac{(n+1)v}{2L}-\dfrac{nv}{2L}$

$f_{n+1}-f_{n}=\dfrac{v}{2L}$

$L=\dfrac{v}{2(f_{n+1}-f_{n})}$

Put the value into the formula

$L=\dfrac{372}{2(642-535)}$

$L=1.73\ m$

Hence, The length of the string is 1.73 m.

Answer 2

The length of this string is equal to 1.74 meters.

Given the following data:

• First frequency = 535 Hz.

• Second frequency = 642 Hz.

• Speed = 372 m/s.

How to calculate the length of a string.

In order to determine the length of this string, we would apply the nth and (n+1) harmonic equation as follows:

The nth harmonic equation is given by:

$F_n = \frac{nV}{2L}$   ....equation 1.

The (n+1) harmonic equation is given by:

$F_{n+1} = \frac{(n+1)V}{2L}$

Subtracting eqn. 1 from eqn. 2, we have:

$F_{n+1} -F_n= \frac{(n+1)V}{2L}-\frac{nV}{2L}\\\\F_{n+1} -F_n=\frac{V}{2L}\\\\2L(F_{n+1} -F_n)=V\\\\L=\frac{V}{2(F_{n+1} -F_n)}$

Substituting the given parameters into the formula, we have;

$L=\frac{372}{2(642 -535)}\\\\L=\frac{372}{2(107)}$

L = 1.74 meters.

Read more on frequency here: brainly.com/question/3841958