Q36E
Question
Adjacent antinodes of a standing wave on a string are \(15\;{\rm{cm}}\) apart. A particle at an antinode oscillates in simple harmonic motion with amplitude \(0.850\;{\rm{cm}}\) and period \(0.0750\;{\rm{s}}\). 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 wavelength, 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?
Step-by-Step Solution
Verified(a) The adjacent nodes distance is, \(15\;{\rm{cm}}\).
The given data can be listed below as,
- The distance of antinodes is, \(d = 15\;{\rm{cm}}\).
An antinode, or location where the standing wave's amplitude is at its greatest, is the opposite of a node. These are situated in the middle of the nodes. The amplitude is at its greatest at antinodes.
The adjacent nodes distance is expressed as,
\(\frac{\lambda}{2}=d\) …(1)
Here \(\lambda \) is the wavelength.
Substitute the value of \(d\) in the equation (1).
\(\frac{\lambda }{2} = 15\;{\rm{cm}}\)
Hence the adjacent nodes distance is, \(15\;{\rm{cm}}\).