One end of a horizontal string is attached to a vibrating blade, and the other e

Question

One end of a horizontal string is attached to a vibrating blade, and the other end passes over a pulley as shown in the figure below. A sphere of mass m=2kg hangs on the end of the string. The string is vibrating in its second harmonic. A container of water is raised under the sphere so that the sphere is completely submerged. In this configuration, the string vibrates in its fifth harmonic. What is the radius of the sphere?

One end of a horizontal string is attached to a vibrating blade, and the other end passes over a pulley as shown in the figure below. A sphere of mass m=2kg hangs on the end of the string. The string is vibrating in its second harmonic. A container of water is raised under the sphere so that the sphere is completely submerged. In this configuration, the string vibrates in its fifth harmonic. What is the radius of the sphere?

Explanation / Answer

In first case

Tension in the string

T1 = m*g = 20 N

In second case

Tension in the string

T2 + Fb = m*g = 20

Fb is Buyoncy force

Fb = rho*V*g = 1000*4/3*Pi*r^3*10

Fb = 4.19*10^4*r^3---------(1)

But we know

Fundamental frquency,fo =c/lambda---------(2)

and

c = Sqrt(T/p)

p is linear desity of string

For case 1

c1 = Sqrt(T1/p)

For case 2

c2 = Sqrt(T2/p)

Let L be the length of the string

lambda =2*L

f1 = c1/lambda

f2 = c2/(lambda)

f1/f2 = 5/2

2*c1 = 5*c2

25*T1 = 4*T2

T2 = 4/25*T1 = 3.2 N

Fb = 20 -3.2 = 16.8

4.19*10^4*r^3 = 16.8

r = 7.4 cm

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