The vibrations of a string fixed at both ends are described by the equation `y= (5.00 mm) sin [(1.57cm^(-1))x] sin [(314 s^(-1))t]`
(a) What is the maximum displacement of particle at x = 5.66 cm ? (b) What are the wavelengths and the wave speeds of the two transvers waves that combine to give the above vibration ? (c ) What is the velocity of the particle at x = 5.66 cm at time t = 2.00s ? (d) If the length of the string is 10.0 cm, locate the nodes and teh antinodes. How many loops are formed in the vibration ?

Options

A. 2
B. 3
C. 4
D. 5

Answer

D

Explanation

This is an equation of a stationary wave. The coefficient of x represent k.
`impliesk=(2pi)/(lambda)=1.57orlambda=(2(3.14))/(1.57)=4cm`
Since each loop in a stationary wave exists between consecutive nodes, the length of the loop is `(lambda)/(2)`or 2 cm.
`l=n((lambda)/(2))orn=(2l)/(lambda)=(2(10cm))/(4cm)=5`