A transverse wave on a string is described by the following wave function. y = 0

Question

A transverse wave on a string is described by the following wave function.

y = 0.085 sin

x + 3?t

where x and y are in meters and t is in seconds.

(a) Determine the transverse speed at t = 0.240 s for an element of the string located at x = 1.50 m.
m/s

(b) Determine the transverse acceleration at t = 0.240 s for an element of the string located at x = 1.50 m.
m/s2

(c) What is the wavelength of this wave?
m

(d) What is the period of this wave?
s

(e) What is the speed of propagation of this wave?
m/s

Explanation / Answer

general equation of wave on a string is given by

Y(x,t) = A sin(kx+ wt)

where A is amplitude

k is wave no. = 2pi/L , L being wave length

W is angular frequency = 2pif , f being frequency

so now given wave equation is Y = 0.085 sin (11x + 3pi t)

we have now by comparision,
------------------------------------
part A :

v = dy/dt = -0.085 * 3pi * cos ( 11x +3pi t)

v at x = 1.5 and t = 0.24 secs is

v = -0.085 * 3*3.14   * cos ( (11* 1.5) + ( 3*3.14 *0.24)

v = - 0.797 m/s

-----------------------------------------

part B :

accleration a = dv./dt

a = 9pi^2 * 0.085 sin (11 x + 3pit)

at x = 1.5 m and t = 0.24 s

a = 9*3.14 * 3.14 * 0.085 * sin ((11* 1.5) + ( 3*3.14* 0.24))

a = -1.66 m/s^2

-------------------------

part C:

K = 2pi/L = 11

wavelength L =- 2pi/11

L = 0.57 m
-----------------------

part D :

Wt = 3pi t

W =2pi f =    3pi

f = 3/2

t = 1/f = 2/3 = 0.667 secs

--------------------------

part E:

use V = Lf

so speed V = 0.57/0.667

V = 0.854 m/s

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