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

boitshupoO

boitshupoO

Answered question

2021-02-14

In the arrangement shown in Figure P14.40, an object of mass,m=2.0kg, hangs from a cordaround a light pulley. The length of the cord between pointP and the pulley is L=2.0m.

(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 (if any) 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?

Answer & Explanation

pattererX

pattererX

Skilled2021-02-16Added 95 answers

mass m=2kg
length of the cord L=2m
(a) frequency in above situation F=145Hz
F=6 fundamental frequency since we have sixloops
=6f
i.e.,6f=145Hz
f=24.16667Hz
fundamental frequency f=24.16667Hz
(12L)Tμ=24.16667Hz
where T= tension in the string =mg
substitute values we get linear density μ value
( b ) .fundamental frequency f=(12L)Tμ
=(12L)mgμ
f1f2=m1m2
we know f1=24.16667 Hz
m1=2 kg
m2=2.88
from above equation we find f 2 value
( c) .similar procedure as problem ( b )

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