Wave Pulse A wave pulse travels down a slinky. The mass of the slinky is m 0.91

Question

Wave Pulse A wave pulse travels down a slinky. The mass of the slinky is m 0.91 kg and is initially stretched to a length L- 6.6m. The wave pulse has an amplitude of A 0.24 m and takes t 0.456 s to travel down the stretched length of the slinky. The frequency of the wave pulse is f-0.44 Hz. What is the speed of the wave pulse What is the tension in the slinky? What is the average speed of a piece of the slinky as a complete wave pulse passes? What is the wavelength of the wave pulse? O5 Now the slinky is stretched to twice its length (but the total mass does not change). What is the new tension in the slinky? (assume the slinky acts as a spring that obeys Hooke's Law) OWhat is the new mass density of the slinky? ?7) what is the new time it takes for a wave pulse to travel down the slinky? OIf the new wave pulse has the same frequency, what is the new wavelength? o" What does the energy of the wave pulse depend on? Othe frequency Othe amplitude Oboth the frequency and the amplitude

Explanation / Answer

1) wave speed, v = L/t

= 6.6/0.456

= 14.5 m/s <<<<<<<<--------------Answer

2) linear mass density of the slinky, mue = m/L

= 0.91/6.6

= 0.138 kg/m

we know, v = sqrt(T/mue)

v^2 = T/mue

==> T = v^2*mue

= 14.5^2*0.138

= 29.0 N <<<<<<<<<--------------Answer

3) V_avg = distance traveled/time taken

= 4*A/T

= 4*A*f

= 4*0.24*0.44

= 0.422 m/s <<<<<<<<<--------------Answer

4) lamda = v/f

= 14.5/0.44

= 32.9 m

5) T = 2*29

= 58.0 N

6) mue = m/(2*L)

= 0.91/(2*6.6)

= 0.0689 kg/m

7) t = v/L

= sqrt(T'/mue')/L

= sqrt(58/0.0689)/6.6

= 4.40 s

8) v' = 4*v = 4*14.4 = 57.6 m/s

lamda' = v'/f

= 57.6/0.44

= 131 m

9) Both the frequency and the amplitude

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