Question

1. (a) If y1 = (2.50 cm)sin [(3.25 cm−1)x − (1.85 s−1)t] and y2 = (3.25...

1. (a) If y1 = (2.50 cm)sin [(3.25 cm−1)x − (1.85 s−1)t] and y2 = (3.25 cm)sin [(2.15 cm−1)x + (1.25 s−1)t] cm represent two waves moving toward each other on a string, find the superposition of the two waves at x = 1.45 cm and t = 2.35 s.(2.15 cm−1)x + (1.25 s−1)t

-cm

(b) Now consider the same two waves on the same string. If they are both moving to the left instead of opposing directions, find the superposition of the waves at x = 1.45 cm and t = 2.35 s. (Assume the +x direction is to the right.)

-cm

2. At t = 0, the instantaneous position of two pulses moving along a taut string with a speed v = 2.93 cm/s are as shown in the diagram below. Each unit on the horizontal axis is 2.0 cm and each unit on the vertical axis is 8.0 cm.

(a) At what location will the resultant of the two pulses have minimum amplitude?
cm

(b) At what time will the resultant of the two pulses have minimum amplitude?
s

(c) What is the value of this minimum amplitude?
cm

0 0
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Answer #1

ANSWER :1) solf 0 (al y = 4 + ₂ 99.5 xỉn ( 3.25 -1.35+ + 3.2 5 3) (2-151+ 1.951) y is resultant wave of super position of y, and yeat x = 1:45, t=2.35 y = 2.5 sin (0.365 )+ 3,255in (0.18) y = 1.47 cm (9) Solo tere, given V=2093cmis. horizontal axis is 2.0cORO distance = 10x2=20cm time to meet zoom. = 3.413 S. 5-86cms (6) (a) minimum amplitude is Kompo = 8+. (hold) 22.93) 93) — 1

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