Question

Cousin Throckmorton is playing with the clothesline. One end of the clothesline is attached to a vertical post.

 

Cousin Throckmorton is playing with the clothesline. One end of the clothesline is attached to a vertical post. Throcky holds the other end loosely in his hand, so that the speed of waves on the clothesline is a relatively slow 0.740m/s. He finds several frequencies at which he can oscillate his end of the clothesline so that a light clothespin 43.0cm from the post doesn't move.

 

What are these frequencies? 

 (0.860 kHz)n, n=1, 2, 3...

 (0.860Hz )n, n=1, 2, 3...

 (1.72 kHz)n, n= 1, 2, 3...

 (1.72Hz )n, n= 1, 2, 3...


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Answer #1
Concepts and reason

Use the concepts of vibrations in a stretched string to solve the given problem.

Consider the vibrations in a stretched string and find the fundamental frequency using the frequency formula. And then, find the few other frequencies (first overtone and second overtone, etc.) by multiplying the fundamental frequency with number of loops.

Fundamentals

The frequency ff expression is,

f=vλf = \frac{v}{\lambda }

Here, λ\lambda is the wavelength and vv is the speed of the wave.

When the string vibrates with fundamental frequency, the string forms only one loop.

So, the length L of the string between the two fixed points is,

L=λ2λ=2L\begin{array}{c}\\L = \frac{\lambda }{2}\\\\\lambda = 2L\\\end{array}

The fundamental frequency is,

f1=vλ{f_1} = \frac{v}{\lambda }

Replace λ\lambda with 2L2L .

f1=v2L{f_1} = \frac{v}{{2L}}

When the two loops are formed between the two fixed points in the stretched string, then the length of the string is,

L=(2)λ2λ=L\begin{array}{c}\\L = \left( 2 \right)\frac{\lambda }{2}\\\\\lambda = L\\\end{array}

The first overtone frequency is,

f2=vλ{f_2} = \frac{v}{\lambda }

Replace λ\lambda with LL .

f2=vLf2=(2)v2L\begin{array}{l}\\{f_2} = \frac{v}{L}\\\\{f_2} = \left( 2 \right)\frac{v}{{2L}}\\\end{array}

Similarly, when the string vibrates with three loops, then

L=(3)λ2λ=2L3\begin{array}{c}\\L = \left( 3 \right)\frac{\lambda }{2}\\\\\lambda = \frac{{2L}}{3}\\\end{array}

The second overtone frequency is,

f3=vλ{f_3} = \frac{v}{\lambda }

Replace λ\lambda with 2L3\frac{{2L}}{3} .

f3=v2L3f3=(3)v2L\begin{array}{l}\\{f_3} = \frac{v}{{\frac{{2L}}{3}}}\\\\{f_3} = \left( 3 \right)\frac{v}{{2L}}\\\end{array}

Thus, from the above discussion, it can be concluded that the harmonic frequencies of the vibrating string can be expressed as integral nn multiple of fundamental frequency. Thus,

f=nv2Lf = n\frac{v}{{2L}}

Substitute 0.740 m/s for v and 0.430 m for L.

f=n0.740m/s2(0.430m)f=n(0.86Hz)\begin{array}{c}\\f = n\frac{{0.740{\rm{ m/s}}}}{{2\left( {0.430{\rm{ m}}} \right)}}\\\\f = n\left( {0.86{\rm{ Hz}}} \right)\\\end{array}

Here, n is equal to integer values from 1, 2, 3, ……n.

Ans:

Thus, the frequencies observed are n(0.86Hz)n=1,2,3,....n\left( {0.86{\rm{ Hz}}} \right){\rm{ }}n = 1,2,3,.... .

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