A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equationy(x, t)=(8.50mm)cos(172rad⋅m−1x−2730rad⋅ s−1t)Assume that the tension of the string is constant and equal to W. How much time does it take a pulse to travel the full length of the string?What was the weight W?How many wavelengths are on the string at any instant of time?What is the equation for waves traveling down the string?

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

The relationship between velocity and propagation constant is

v = 15.87m/s

Time taken,

t = 0.0819s

2)

The velocity of transverse wave is given by

mass of string is calculated thus

mg = 0.0125N

m = 0.00128kg

0.25N

3)

The propagation constant k is

hence

0.036 m

No of wavelengths, n is

n = 36

4)

The equation of wave travelling down the string is

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.