3-1 In Fig. 15-31, two identical springs of spring constant 7580 N/m are attached to ablock of mass 0.245 kg. What is the frequency of oscillation on the frictionless floor? (HR 15-13)
Sol: When displaced from equilibrium, the net force exerted by the springs is –2kx acting in a direction so as to return the block to its equilibrium position (x = 0). Since the acceleration  , Newton’s second law yields
Substituting x = xm cos(t + ) and simplifying, we find
where is in radians per unit time. Since there are 2 radians in a cycle, and frequency f measures cycles per second, we obtain
3-2 In Fig. 15-36, two springs are joined and connected to a block of mass 0.245 kg that is set oscillating over a frictionless floor. The springs each have spring constant k = 6430 N/m. What is the frequency of the oscillations? (HR 15-26)
Sol: We wish to find the effective spring constant for the combination of springs shown in the figure. We do this by finding the magnitude F of the force exerted on the mass when the total elongation of the springs is x. Then keff = F/x. Suppose the left-hand spring is elongated by  and the right-hand spring is elongated by xr. The left-hand spring exerts a force of magnitude  on the right-hand spring and the right-hand spring exerts a force of magnitude kxr on the left-hand spring. By Newton’s third law these must be equal, so  . The two elongations must be the same and the total elongation is twice the elongation of either spring:  . The left-hand spring exerts a force on the block and its magnitude is  . Thus  . The block behaves as if it were subject to the force of a single spring, with spring constant k/2. To find the frequency of its motion replace keff in  with k/2 to obtain
With m = 0.245 kg and k = 6430 N/m, the frequency is f = 18.2 Hz.
3-3 For Eq. 15-45, suppose the amplitude xm is given by
where Fm is the (constant) amplitude of the external oscillating force exerted on the spring by a rigid support in Fig. 15-15. At resonance, what are the (a) amplitude and (b) velocity amplitude of the oscillating object? (HR 15-61)
Sol: (a) We set = d and find that the given expression reduces to xm = Fm/b at resonance.
(b) since the velocity amplitude vm = xm, at resonance, we have vm = Fm/b = Fm/b.
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