13) The frequency of a vibrating string varies inversely as its length. A string 3 feet long vibrates 175 cycles per second. Find the frequency of a 5 foot string. 14) The force of the wind blowing on a vertical surface varies jointly as the area of the surface and the square of the velocity. If a wind blowing at 50 mph exerts a force of 75 pounds on a surface of 500 ft2 , how much force will a wind of 75 mph place on a surface of 10 ft2?

13) f2 = 105 cycles/sec

14) F2 = 3.375 lbs

Explanation:

Part 13)

- The frequency (f) of a vibrating string varies inversely as its length (L):

                                   f = k / L

Where,                        k = proportionality constant

- Set up proportionality constant to equations as follows:

                                   k = f1*L1 = f2*L2

Where,                        f1 = 175 cycles/s , L1 = 3 ft

                                   f2 = ? , L2 = 5 ft

- Using the relationship we determine f2:

                                  f2 = f1*L1 / L2

                                  f2 = 175*3/5

                                  f2 = 105 cycles/sec

Part 14)

- The force (F) of the wind blowing on a vertical surface varies jointly as the area of the surface (A) and the square of the velocity ( v^2):

                                  F = k*A*v^2

                                  F / A*v^2 = k

Where,                       k = proportionality constant

- Set up proportionality constant to equations as follows:

                                  F1 / A1*v1^2 =  F2 / A2*v2^2  

Where,                       F1 = 75 lbs , A1 = 500 ft^2 , v = 50 mph

                                  F2 = ? , A2 = 10 ft^2 , v = 75 mph

- Using the relationship we determine F2:

                                  F2 = F1 * ( A2*v2^2 / A1*v1^2 )

                                  F2 = 75*( 10*75^2 / 500*50^2 )

                                  F2 = 3.375 lbs

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