Waves and strings?

In figure 16-38a, string 1 has a linear density of 3.70 g/m, and string 2 has a linear density of 7.10 g/m. They are under tension due to the hanging block of mass M = 351 g. Calculate the wave speed on (a) string 1 and (b) string 2. (Hint: When a string loops halfway around a pulley, it pulls on the pulley with a net force that is twice the tension in the string.) Next the block is divided into two blocks (with M1 + M2 = M) and the apparatus is rearranged as shown in figure 16-38b. Find (c) M1 and (d) M2 (in grams) such that the wave speeds in the two strings are equal.


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Waves and strings?
d1 = 3.70 g/m = 3.70x10^-3 kg/m


d2 = 7.10 g/m = 7.10x10^-3 kg/m


M = 351g = 0.351kg


the total tension is Mg and this is shared between both strings evenly so there is a tension of


T = Mg/2


in each string


then


v = √(T/d) = √(Mg/(2d))


plug d1 and d2 in this to find v1 and v2 respectively, gives


a) v1 = 21.6m/s


b) v2 = 15.6m/s





M = M1 + M2


the tension in each string is


T1 = M1g %26amp; T2 = M2g


in this case you have


v1 = √(T1/d1) = √(M1g/d1)


%26amp;


v2 = √(T2/d2) = √(M2g/d2)


so for the velocities to be equal you need


√(M1g/d1) = √(M2g/d2)


M1/d1 = M2/d2


so substitute M2 = M1d2/d1 in


M = M1 + M2 = M1( 1 + d2/d1)


gives


c) M1 = M / ( 1 + d2/d1) = Md1/(d1 + d2) ≈ 0.120kg = 120g


substitute this in


M2 = M1d2/d1


gives


d) M2 = Md2/(d1 + d2) ≈ 0.231kg = 231g





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