1 ) A horizontal segment of pipe tapers from a cross sectional area of 50 cm^2 t

Question

1 ) A horizontal segment of pipe tapers from a cross sectional area of 50 cm^2 to 0.5 cm^2. The pressure at the larger end of the pipe is 1.19 10^5 Pa and the speed is 0.027 m/s. What is the pressure at the narrow end of the segment?

2 ) The prong of a tuning fork moves back and forth when it is set into vibration. The distance the prong moves between its extreme positions is 2.21 mm. If the frequency of the tuning fork is 440.5 Hz, what are the maximum velocity and the maximum acceleration of the prong?
vm = 1 m/s
am = 2 m/s2

3 ) An empty cart, tied between two ideal springs, oscillates with ? = 11.0 rad/s (the figure below). A load is placed in the cart, making the total mass 3.9 times what it was before. What is the new value of ??

Explanation / Answer

1) the areas of the pipe , A1 = 50 cm2 and A2 = 0.5 cm2 the pressure P1 = 1.19 105 Pa speed v1 = 0.027 m/s density = 1000 kg/m3 using , equation of continuity ,             A1v1 = A2v2 so , speed v2 = (A1/A2)v1                      = (50/0.5)(0.027)                      = 2.7 m/s apply Bernoulli's theorem ,     P1 + 1/2 v12 = P2 + 1/2 v22    P2 = P1 + 1/2 [v12 - v22]         = 1.19 105 + (1/2)(1000)[0.0272 - 2.72]         =    119000 - 3644.6355         = 1.153*105 Pa ......................................................................... 2) Frequency   f = 440.5 Hz
Amplitude A = 2.21 mm/2
                    = 1.105 mm
                    = 1.105*10-3 m
the max velocity is
           vmax = A
                    = A(2f)                     = (1.105*10-3)(2)(3.14)(440.5)                     = 3.056 m/s
the max acceleration is
         amax = A2
                 = A(2f)2                  = (1.105*10-3)(2*3.14*440.5)2                  = 8456.16 m/s2 ....................................................... 3) angular speed 1 = 11 rad/s Time period is                 T= 2[m/(k1+k2)]
angular speed is                   = 2/T                      = [(k1+k2)/m]      .............. (1) therefore ,         1 = [(k1+k2)/m1]     ............ (2)         2 = [(k1+k2)/m2]    .......... (3) but mass m2 = (3.9)m1 equation (3) , becomes        2 = [(k1+k2)/(3.9)m1]   .............. (4)      
      1/2 = (3.9)                  = (1.9748)     2 = 1/1.9748           = 11 / 1.9748           = 5.57 rad/s
....................................................... 3) angular speed 1 = 11 rad/s Time period is                 T= 2[m/(k1+k2)]
angular speed is                   = 2/T                      = [(k1+k2)/m]      .............. (1) therefore ,         1 = [(k1+k2)/m1]     ............ (2)         2 = [(k1+k2)/m2]    .......... (3) but mass m2 = (3.9)m1 equation (3) , becomes        2 = [(k1+k2)/(3.9)m1]   .............. (4)      
      1/2 = (3.9)                  = (1.9748)     2 = 1/1.9748           = 11 / 1.9748           = 5.57 rad/s

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