One end of a horizontal string is attached to a moveable support, while the othe

Question

One end of a horizontal string is attached to a moveable support, while the other passes over a frictionless pulley and is tied to a 1.5 kg sphere. Initially, the sphere is hanging in the air over a bucket of water and the frequencies of standing waves in the horizontal portion of the string are measured. The support is then slid half the distance to the pulley, completely submerging the sphere in the water. The frequencies of standing waves are again measured, and it's found that the fourth harmonic (m -4) frequency when the sphere is completely submerged exactly matches the fifth harmonic (m 5) frequency from before it was submerged What is the diameter of the sphere? Initial Final L/2

Explanation / Answer

speed(V) = sqrt(T*l/m) T is tension, l is length of string, m is mass of string

Frequency = V/lambda lambda is wavelength

lambda= 2*L/N N is Nth harmonic

=> F = sqrt(T*L/m)/(2*l/N) => F^2 = T*L*N^2/m*4*L^2 = T*N^2/4mL

=> F1^2 = F2^2

T2 = mg = 9.8*1.5 = 14.7 N

=> T1*4*4/4*m*(L/2) = T2*5*5/4*m*L

=> T1 = 25*T2/32 = 25*9.8*1.5/32 = 11.484

=> buoyant force = T2 - T1 = 14.7 - 11.484 = 3.216 = volume*density of water

=> V = buoyant force/density = 3.216/(1000) = 0.003216 = 4*pi*r^3/3

=> r = (3*0.003216/(4*pi))^(1/3) = 0.0915 m

=> diameter = 2*r = 2*0.0915 = 0.183 m Answer

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