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

i think the answer is a, hope this is right, that would be my best guess.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s