Question

A sinusoidal transverse wave travels along a long, stretched string. The amplitude of this wave is 0.0863 m, its frequency is 2.89 Hz, and its wavelength is 1.13 m. What is the shortest transverse distance d between a maximum and a minimum of the wave? d = m How much time At is required for 75.7 cycles of the wave to pass a stationary observer? At = S Viewing the whole wave at any instant, how many cycles N are there in a 33.5 m length of string? N= cycles

Answer 1

Given that Amplitude, A=0.0863m frequency, f = 2.89 Hz wavelength, is = 113m * Shortest trasverse distance &ľ between maximum and minimum of the wave 2 x amplitude of wave d = 2 X A - 2x0.0863 d = 0.1726 m Am * frequency f = 2,8972 2.89 cycersee stationary I see 2.89 cycle pas a observer in 1 cque pars a stationary observer in in I see 2.89 75.7 cycles pars a stal stalxonary observer in At = LX 75.7 2.89 at = 26.2 seconds

* N = Total length of string wave longty 33.5 I 1:13 르 29.6 cycles