Adjacent antinodes of a standing wave on a string are 15.0 cm apart. A particle

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 is the wavelength of the two traveling waves that form this pattern?

(c) What is the amplitude of the two traveling waves that form this pattern?

(d) What is the speed of the two traveling waves that form this pattern?

(e) Find the maximum and minimum transverse speeds of a point at an antinode.

(f) What is the shortest distance along the string between a node and an antinode?

Explanation / Answer

(a) The nodes are likely to be the same distance apart as the antinodes =15.0 cm

(b) The nodes are half a wavelength apart, so the wavelength is just twice the distance between the nodes =2*15.0=30.0cm

(c) The amplitude of the travelling waves is combining destructively at the nodes, giving a zero net amplitude, and constructively at the antinodes, giving an amplitude twice that of each individual travelling wave (assuming they have the same amplitude, which you'll have to). So the amplitude of the travelling waves will be half that of the standing wave at the antinodes =0.5*0.85=0.425 cm

(d) You have a period, which you can do 1/period to get frequency, and a wavelength, so to find the speed you can use v = f(lambda) where f is the frequency and lambda is the wavelength.

v = f*lambda =0.3/0.0750= 4 m/s.

(e) The minimum transverse speed at the antinodes will be zero: the string has zero transverse speed at its maximum displacement. Its maximum transverse speed can be calculated using what you know about speeds in simple harmonic motion (SHM) combined with the period and the amplitude. It will occur as the string passes through the middle of its motion.

Transverse v = 2pi*f*amplitude = 2*pi*13.33*0.00425=0.356 m/s.

(f) The node-to-antinode distance will just be half the node-to-node or antinode-to-antinode distance =15/2 =7.5 cm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.