Question

A string vibrates according to the equation y(x,t) = 2.0sin(0.16x)cos(750t), where x and y are in centimeters and t is in seconds.

(a) What are the amplitude and velocity of the component waves whose superposition give rise to this vibration?

(b) What is the distance between nodes?

(c) What is the velocity of a particle of the string at the position x = 9.0 cm when t=5x10-3

Answer 1

Answer with Explanation:

We are given that an equation

$y(x,t)=2.0sin(0.16x)cos(750t)$

Where x and y are in cm and t is in sec.

a.Compare with

$y(x,t)=Asin(kx)cos(\omega t)$

Where Amplitude=$\frac{A}{2}$

Amplitude=$\frac{2}{2}=1 cm$

k=0.16 and $\omega=750 rad/s$

Velocity=$\frac{\omega}{k}$

Velocity=$\frac{750}{0.16}=4687.5cm/s=\frac{45687.5}{100}m/s=46.875m/s$

1 m=100 cm

Velocity of the component wave=46.875m/s

b.The distance between two nodes=$\frac{\lambda}{2}$

$\lambda=\frac{2\pi}{k}=\frac{2\times 3.14}{0.16}=39.3cm$

Therefore, distance between two nodes=$\frac{39.3}{2}=19.65 cm$

c.x=9 cm,

t=$5\times 10^{-3}s$

$v(x,t)=\frac{dy}{dt}$

Using the formula

$v(x,t)=2sin(0.16x)\times (-750sin(750t))$

$v(x,t)=-1500sin(0.16x)sin(750t)$

Substitute the values

$v(x,t)=-1500sin(0.16\times 9)sin(750\times 5\times 10^{-3})$

$v(x,t)=850cm/s=\frac{850}{100}m/s$

$v(x,t)=8.5m/s$