A transverse wave on a string has an amplitude of 15cm, a wavelenght of 25 cm, and a frequencyof 2.0 Hz. 1) What is the mathematical description of the displacement from equilibrium for the wave at t = 0, x = 0 and y = 0? a) y(x,t) = (0.15m) sin ((25.13 m-1)x - (12.57 rad/s) t) b) y(x,t) = (0.15m) cos ((25.13 m-1)x - (12.57 rad/s) t) c) y(x,t) = (0.15m) sin ((15710 m-1)x - (12.57 rad/s) t) d)  y(x,t) = (0.15m) cos ((15710 m-1)x - (12.57 rad/s) t)   2) What is the mathematical description of the displacement from equilibrium for the wave at t = 0, x = 0 and y = +15cm? a) y(x,t) = (0.15m) sin ((25.13 m-1)x - (12.57 rad/s) t) b) y(x,t) = (0.15m) cos ((25.13 m-1)x - (12.57 rad/s) t) c) y(x,t) = (0.15m) sin ((15710 m-1)x - (12.57 rad/s) t) d)  y(x,t) = (0.15m) cos ((15710 m-1)x - (12.57 rad/s) t)   3) What is the mathematical description of the displacement from equilibrium for the wave at t = 0, x = 0 and y = -15cm? a) y(x,t) = -(0.15m) sin ((25.13 m-1)x - (12.57 rad/s) t) b) y(x,t) = -(0.15m) cos ((25.13 m-1)x - (12.57 rad/s) t) c) y(x,t) = -(0.15m) sin ((15710 m-1)x - (12.57 rad/s) t) d)  y(x,t) = -(0.15m) cos ((15710 m-1)x - (12.57 rad/s) t)   4) What is the mathematical description of the displacement from equilibrium for the wave at t = 0, x = 0 and y =12 cm?  a) y(x,t) = (0.15m) sin ((25.13 m-1)x - (12.57 rad/s) t + 0.644) b) y(x,t) = (0.15m) cos ((25.13 m-1)x - (12.57 rad/s) t + 0.644) c) y(x,t) = (0.15m) sin ((15710 m-1)x - (12.57 rad/s) t + 0.644) d) y(x,t) = (0.15m) cos ((15710 m-1)x - (12.57 rad/s) t  + 0.644)

Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter13: Mechanical Waves
Section: Chapter Questions
Problem 10P: A transverse wave on a string is described by the wave function y=0.120sin(8x+4t) where x and y are...
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A transverse wave on a string has an amplitude of 15cm, a wavelenght of 25 cm, and a frequencyof 2.0 Hz.

1) What is the mathematical description of the displacement from equilibrium for the wave at t = 0, x = 0 and y = 0?

a) y(x,t) = (0.15m) sin ((25.13 m-1)x - (12.57 rad/s) t)

b) y(x,t) = (0.15m) cos ((25.13 m-1)x - (12.57 rad/s) t)

c) y(x,t) = (0.15m) sin ((15710 m-1)x - (12.57 rad/s) t)

d)  y(x,t) = (0.15m) cos ((15710 m-1)x - (12.57 rad/s) t)

 

2) What is the mathematical description of the displacement from equilibrium for the wave at t = 0, x = 0 and y = +15cm?

a) y(x,t) = (0.15m) sin ((25.13 m-1)x - (12.57 rad/s) t)

b) y(x,t) = (0.15m) cos ((25.13 m-1)x - (12.57 rad/s) t)

c) y(x,t) = (0.15m) sin ((15710 m-1)x - (12.57 rad/s) t)

d)  y(x,t) = (0.15m) cos ((15710 m-1)x - (12.57 rad/s) t)

 

3) What is the mathematical description of the displacement from equilibrium for the wave at t = 0, x = 0 and y = -15cm?

a) y(x,t) = -(0.15m) sin ((25.13 m-1)x - (12.57 rad/s) t)

b) y(x,t) = -(0.15m) cos ((25.13 m-1)x - (12.57 rad/s) t)

c) y(x,t) = -(0.15m) sin ((15710 m-1)x - (12.57 rad/s) t)

d)  y(x,t) = -(0.15m) cos ((15710 m-1)x - (12.57 rad/s) t)

 

4) What is the mathematical description of the displacement from equilibrium for the wave at t = 0, x = 0 and

y =12 cm? 

a) y(x,t) = (0.15m) sin ((25.13 m-1)x - (12.57 rad/s) t + 0.644)

b) y(x,t) = (0.15m) cos ((25.13 m-1)x - (12.57 rad/s) t + 0.644)

c) y(x,t) = (0.15m) sin ((15710 m-1)x - (12.57 rad/s) t + 0.644)

d) y(x,t) = (0.15m) cos ((15710 m-1)x - (12.57 rad/s) t  + 0.644)

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