physic EXAM 2 PHY 206 1. n the arrangement shown in Fig, a mass can be hung from
ID: 1470332 • Letter: P
Question
physic EXAM 2 PHY 206 1. n the arrangement shown in Fig, a mass can be hung from a string (with a linear mass density of -0.002kg/m) that passes oer a light pulley. The string is connected to a vitnaar (of costant frequency and the length of the string between point Pand the pulley isL = 2 m, when de nass is eidher 16.0 kg or 250 kg, standing waves are observed: however, so standing waves ane ed; however, so standing waves are observed with any mass between these valoes (a) What is the frequency of the vibralor? (b) What is the largest mass for which standing waves could be observed? (Hint: The greater the tension in the string, the smaller the number of nodes in the standing wave) b) What i the larps mass for ahich stadling waves Vibrator PulleyExplanation / Answer
a) L=2m, µ=0.002kg/m , m1= 25kg , m2= 16kg
v= sqrt(T/µ) --------------(1)
Use equation
f= nv/2L-----------------(2)
From (1)
f= n/2L* sqrt(T/µ) -------------(3)
Assume for m1=25kg , n=n
Hence
f= n/2L* sqrt(T1/µ) -------------(4)
Assume for m1=16kg , n=n+1
Hence
f= (n+1)/2L* sqrt(T2/µ) -------------(5)
Equating (4) and (5)
n/2L* sqrt(T1/µ) = (n+1)/2L* sqrt(T2/µ)
n+1/n = sqrt(T1/T2)
n+1/n = sqrt(m2g/m1g) = sqrt(25/16) = 5/4
gives , n= 4
Plugging in (4)
f= 4/(2*2)*sqrt[(25*9.8)/0.002] = 350 Hz
b) For largest mass n=1 ,
Put this in (4)
350= 1/(2*2)*sqrt[(m*9.8)/0.002] => m = 400kg
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