The Effective Force Constant of Two Springs. Two springs with the same unstretched length but different force constants k1 and k2 are attached to a block with mass m on a level, frictionless surface. Calculate the effective force constant keff in each of the three cases (a), (b), and (c) depicted in ?Fig. P14.92?. (The effective force constant is defined by ) (d) An object with mass ?m?, suspended from a uniform spring with a force constant k , vibrates with a frequency f1. When the spring is cut in half and the same object is suspended from one of the halves, the frequency is f2. What is the ratio f1/f2?

Solution 101CP Step 1: Consider the figure a) In first case imagine the mass m moves a distance x and similarly the springs k1 and k2 moves x1 and x2 .with force f1 = -k x1 and f2=-k x2 1 2 x = x1 = x2 F =f1 + f2 The force constant is given by Fx=-k .x eff So F x(k1+k2) dx So k =k1+k2 eff Step 3: b) In this case x = x1 x2 So K eff+k1 2 Step 4: c) in this case F= f =1 2 x = x1 + x2 x = (1/k1 + 1/k2) f So k eff k1×k2/k1+k2) Step 5: Consider a spring hanging a mass m vertically Frequency of vibration of the spring per unit time initially is f1= 1 k/m 2 By cutting the spring into half let us take f be the2requency and substitute 2k for k So f2 2 2k/m Comparing f and1 we c2 find out the ratio of f and f 1 2 F2 2f 1