Answer to Question #288774 in Physics for NICKO

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 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?

Answer 1

Given:

$d=15.0\:\rm cm$

$A=0.850\:\rm cm$

$T=0.0750\:\rm s$

(a) distance between the adjacent nodes is equal to distance between the adjacent antinodes

$$d=15.0\:\rm cm$$

(b) the wavelength of the wave

$$\lambda=2d_1=30.0 \:\rm cm$$

the amplitude of the wave

$$A=0.850\:\rm cm$$

the speed of the wave

$$v=\lambda/T=30.0/0.0750=400\:\rm cm/s=4\: m/s$$

(c) maximum transverse speed of a point at the antinode of the standing wave

$$v_{\max}=A\omega=2\pi A/T\\ =6.28*0.850/0.0750=71.2\: \rm cm/s$$

minimum transverse speed of a point at the antinode of the standing wave

$$v_{\min}=0$$

(d) the shortest distance along the string between a node and an antinode

$$d=\lambda/4=30.0/4=7.5\:\rm cm$$