A swing is hung from the limb of a tree. When a young child gets into the swing,

Question

A swing is hung from the limb of a tree. When a young child gets into the swing, it settles 2.0 cm. Because of the flexibility of the tree limb, the swing bobs up and down as well as swinging back and forth. When the swing is gently bobbed up and down with no swinging motion, the frequency is 10 times the frequency of the swing when it is gently swung so as not to excite bobbing motion. How long is the swing?

(b) The six strings of a classical guitar are all 65.5 cm long and are tuned to frequencies of 82, 110, 147, 196, 247, and 330 Hz. The mass per unit length of the set of strings is, starting with the one of lowest frequency, 5.05 x 10-3, 3.71 x 10-3, 2.21 x 10-3, 1.01 x 10-3, 0.58 x 10-3, and 0.44 x 10-3 kg/m. (i) What is the tension in each string? (ii) What is the total string force?

Explanation / Answer

assuming the swing to be a simple pendulum,

its frequency is given by

f=(1/2*pi)*sqrt(g/l)

where l=length of the swing

now when length is l+2 cm, its frequency is 1/10 times when length is l.

then (1/2*pi)*sqrt(g/(l+2))=0.1*(1/2*pi)*sqrt(g/l)

==>sqrt(g/(l+2))=0.1*sqrt(g/l)

==>g/(l+2)=0.01*g/l

==>l+2=100*l

==>l=2/99=2.02 cm

part b:

as we know, frequency of the string=(1/2*length)*sqrt(tension/mass per unit length)

hence sqrt(tension/mass per unit length)=2*length*frequency

==>tension=mass per unit length*4*length^2*frequency^2

length is given as 0.655 m

mass per unit length frequency tension

5.05*10^(-3) 82 58.2722 N

3.71*10^(-3) 110 77.03 N

2.21*10^(-3) 147 81.953 N

1.01*10^(-3) 196 66.585 N

0.58*10^(-3) 247 60.7245 N

0.44*10^(-3) 330 82.2282 N

total string force=sum of all the tension values=426.7929 N

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