Question

Adjacent antinodes of a standing wave on a string are 15.0 cm apart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.850 cm and period 0.0750 s. The string lies along the +x-axis and is fixed at x=0.
(A) How far apart are the adjacent nodes?
(B) What are the wave-length, amplitude, and speed of the two traveling waves that form this pattern?
(C) Find the maximum and minimum transverse speeds of a point at an antinode.
(D) What is the shortest distance along the string between a node and an antinode?

Answer 1

Answer:

Explanation:

A ) Distance between two adjacent  anti-node will be equal to distance between two adjacent  nodes . So the required distance is 15 cm .

B )  wave-length, amplitude, and speed of the two traveling waves that form this pattern are as follows

wave length = same as wave length of wave pattern formed. so it is 30 cm

amplitude = 1/2 the amplitude of wave pattern formed so it is .850 / 2 = .425 cm

Speed =   frequency x wavelength ( frequency = 1 / time period )

= 1 / .075) x 30 cm

400 cm / m

C ) maximum speed

= ω A

= (2π / T) x A

= 2 X 3.14 x .85 / .075 cm / s

= 71.17 cm / s

minimum speed is zero.

D ) The shortest distance along the string between a node and an antinode

= Wavelength / 4

= 30 / 4

= 7.5 cm