In the arrangement shown in Figure P14.40, an object of mass,m = 2.0 kg, hangs from a cordaround a light pulley. The length of the cord between pointP and the pulley is L = 2.0 m.(a) When thevibrator is set to a frequency of 145Hz, a standing wave with six loops is formed. What must be thelinear mass density of the cord in kg/m? (b) How many loops (ifany) will result if m is changed to 2.88 kg? (c) How many loops (if any) will result if m is changed to72.0 kg?

In the arrangement shown in Figure P14.40, an object of mass,m = 2.0 kg, hangs from a cordaround a light pulley. The length of the cord between pointP and the pulley is L = 2.0 m.(a) When thevibrator is set to a frequency of 145Hz, a standing wave with six loops is formed. What must be thelinear mass density of the cord in kg/m? (b) How many loops (ifany) will result if m is changed to 2.88 kg? (c) How many loops (if any) will result if m is changed to72.0 kg?

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asked 2021-02-14
In the arrangement shown in Figure P14.40, an object of mass,m = 2.0 kg, hangs from a cordaround a light pulley. The length of the cord between pointP and the pulley is L = 2.0 m.(a) When thevibrator is set to a frequency of 145Hz, a standing wave with six loops is formed. What must be thelinear mass density of the cord in kg/m? (b) How many loops (ifany) will result if m is changed to 2.88 kg?
(c) How many loops (if any) will result if m is changed to72.0 kg?

Answers (1)

2021-02-16
mass m=2 kg
length of the cord L = 2 m
(a) frequency in above situation F = 145 Hz
F = 6 * fundamental frequency since we have sixloops
=6f
i.e., 6f = 145 Hz
f = 24.16667 Hz
fundamental frequency f = 24.16667 Hz
\(\displaystyle{\left({\frac{{{1}}}{{{2}{L}}}}\right)}\sqrt{{{\frac{{{T}}}{{\mu}}}}}={24.16667}{H}{z}\)
where T= tension in the string = m g
substitute values we get linear density \(\displaystyle\mu\) value
( b ) .fundamental frequency \(\displaystyle{f}={\left({\frac{{{1}}}{{{2}{L}}}}\right)}\sqrt{{{\frac{{{T}}}{{\mu}}}}}\)
\(\displaystyle={\left({\frac{{{1}}}{{{2}{L}}}}\right)}\sqrt{{{\frac{{{m}{g}}}{{\mu}}}}}\)
\(\displaystyle{\frac{{{f}_{{1}}}}{{{f}_{{2}}}}}=\sqrt{{{\frac{{{m}_{{1}}}}{{{m}_{{2}}}}}}}\)
we know \(\displaystyle{f}_{{1}}={24.16667}\) Hz
\(\displaystyle{m}_{{1}}={2}\) kg
\(\displaystyle{m}_{{2}}={2.88}\)
from above equation we find f 2 value
( c) .similar procedure as problem ( b )
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