Chapter 16 Solutions_8

# Chapter 16 Solutions_8 - Chapter16 3(a The motion from...

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Chapter 16 3. (a) The motion from maximum displacement to zero is one-fourth of a cycle so 0.170 s is one- fourth of a period. The period is T = 4(0.170 s) = 0.680 s. (b) The frequency is the reciprocal of the period: 1 1 1.47Hz. 0.680s f T = = = (c) A sinusoidal wave travels one wavelength in one period: 1.40m 2.06m s. 0.680s v T = = = λ 7. Using v = f λ , we find the length of one cycle of the wave is λ = 350/500 = 0.700 m = 700 mm. From f = 1/ T , we find the time for one cycle of oscillation is T = 1/500 = 2.00 × 10 –3 s = 2.00 ms. (a) A cycle is equivalent to 2 π radians, so that π /3 rad corresponds to one sixth of a cycle. The corresponding length, therefore, is λ /6 = 700/6 = 117 mm. (b) The interval 1.00 ms is half of T and thus corresponds to half of one cycle, or half of 2 π rad. Thus, the phase difference is (1/2)2 π = π rad.

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