In Figure (a), string 1 has a linear density of 3.38 g/m, and string 2 has a lin

Question

In Figure (a), string 1 has a linear density of 3.38 g/m, and string 2 has a linear density of 7.04 g/m. They are under tension due to the hanging block of mass M = 452 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 (b). Find (c) M1 and (d) M2 such that the wave speeds in the two strings are equal.

Explanation / Answer

d1 = 3.38 g/m = 3.38*10^-3 kg/m

d2 = 7.04 g/m = 7.04*10^-3 kg/m

M = 452g = 0.452kg

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 = sqrt(T/d) = sqrt(Mg/(2d))

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

a) v1 = sqrt(0.452*9.8/(2*0.00338)) = 25.6 m/s

b) v2 = sqrt(0.452*9.8/(2*0.00704)) = 17.74 m/s

M = M1 + M2

the tension in each string is

T1 = M1g and T2 = M2g

so we have

v1 = sqrt(T1/d1) = sqrt(M1g/d1)

and

v2 = sqrt(T2/d2) = sqrt(M2g/d2)

so for the velocities to be equal

sqrt(M1g/d1) = sqrt(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) = 0.452*0.00338/(0.00338 + 0.00704) = 0.1466 kg = 146.6 g

and

M2 = M1d2/d1

gives

d) M2 = Md2/(d1 + d2) = 0.452*0.00704/(0.00338 + 0.00704) = 0.3054 kg = 305.4 g

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