A transverse wave is traveling on a string. The displacement y of a particle from its equilibrium position is given by y = (0.021 m)sin(29t - 2x). Note that the phase angle (29t - 2.0x) is in radians, t is in seconds, and x is in meters. The linear density of the string is 1.80 10-2 kg/m. What is the tension in the string?

3.7845 N

Explanation: The wave equation is given by

y = 0.021 sin (29t - 2x)

By comparing this to the general equation of a wave, we have that

y = A sin (ωt - kx)

Where ω = angular frequency.

After comparing, we see that

A = 0.021m, ωt = 29t and - 2x = - kx

Hence ω = 29 rad/s and k = 2.

The velocity of a wave is in relationship with angular frequency and wave number is given as

ω = kv

Where k = wave number

By substituting the parameters, we have that

29 = 2v

v = 29/2 = 14.5 m/s.

The velocity of wave in a spring is calculated using the formulae below.

v = √(T/u)

Where v = linear velocity = 14.5 m/s

T = tension =?

u = linear density = 1.80×10^-2 kg/m

By substituting the parameters, we have that

14.5 = √(T/ 1.80×10^-2)

By squaring both sides, we have that

14.5² = T/ 1.80×10^-2

T = 14.5² × 1.80×10^-2

T = 210.25 × 1.80×10^-2

T = 3.7845 N

hèllo, sorry i really need some more points so if you need this qustion answered than retype it

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west