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One end of a horizontal string is attached to a vibrating blade , and the other end passes over a pulley as in Fig. 7.49 (a). A sphere of mass 2.00 kg hangs at 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 as shown in Fig . 7.49 (b). What is the radius of the sphere ? |
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Answer» Solution :Imagine what happens when the sphere is immersed in the water . The buoyant force acts upward on the sphere , reducing the tension in the string . The change in tension causes a change in the SPEED of waves on the string , which in turn causes a change in the wavelength . This altered wavelength results in the string vibrating in its normal mode rather than the second . The hanging sphere is modelled as a particle in equilibrium . One of the forces acting on it is the buoyant force from the water . we also apply the waves under boundary conditions model to the string . Apply the particle in equilibrium model to the sphere in Fig . 7.49 (a) , identifying ` T_(1)` as the tension in the string as the sphere hangs in air . `T_(1) = MG`(i) Apply the particle in equilibrium model to the sphere in Fig . 7.49 (B) , where `T_(2)` is the tension in the string as the sphere is immersed in water . LTBR. `B + T_(2) = mg rArr B = mg - T_(2)`(ii) The desired quantity , the radius of the sphere , will appear in the expression for the buoyant force `B` . Before proceeding in this direction , however , we must evalute `T_(2)` from the information about the standing wave . WRITE the equation for the frequency of a standing wave on a string twice , once before the sphere is immersed and once after . Notice that the frequency `f` is same in both the cases becauseit is determined by the vibrating blade . In addition , the linear mass density `mu` and the length `L` of the vibrating portion of the string are same in both cases . Divide the two equations. ` f = (n_(1))/( 2L) sqrt((T_(1))/( mu))`(iii) ` f = (n_(2))/( 2L) sqrt ((T_(2))/(mu))`(iv) ` 1 = (n_(1))/( n_(2)) sqrt ((T_(1))/( T_(2)))`(v) solve for `T_(2)`: `T_(2) = ((n_(1))/(n_(2)))^(2) T_(1) = ((2)/(5))^(2) ( 19.6 N) = 3.14 N` Substituting this result into Eq. (ii) ` B = mg - T_(2) = 19.6 N - 3.14 N = 16.5 N` Now`B = rho _( water) g V _(sphere) = rho _(water) g ((4)/(3) pi r^(3))` Solve for the radius of the sphere : ` r = (( 3B)/( 4 pi rho_(water) g ))^(1//3)` ` = ( 3( 16.5 N ))/( 4 pi ( 1000 kg //m^(3)) ( 9.80 m//s^(2)))^(1//3)` ` = 7.38 xx 10^(-2) m = 7.38 cm` Notice that only certain radii of the sphere will result inthe string vibrating in a normal mode , the speed of waves on the string must be changed to a value such that the length of the string is integer multiple of half wavelengths . The radii of the sphere that cause the string to vibrate in a normal mode are quantized. |
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