Two loudspeakers, 5.5 m apart and facing each other, play identical sounds of the same frequency. You stand halfway between them, where there is a maximum of sound intensity. Moving from this point toward one of the speakers, you encounter a minimum of sound intensity when you have moved 0.21 m . Assume the speed of sound is 340 m/s. Part A) What is the frequency of the sound? Part B) If the frequency is then increased while you remain 0.21 m from the center, what is the first frequency for which that location will be a maximum of sound intensity? Express your answer to two significant figures and include the appropriate units. Please show ALL steps, equations AND how you get the frequency in Hz.

a)   f = 809.5 Hz ,  b)    f = 1619 Hz

Explanation:

The condition for destructive interference from the speakers is

             Δr / lam = Ф / 2π

             λ = Δr 2π / π

We substitute

            λ = 0.21 2 π/π

             Lam = 0.42 m

Now let's use the ratio of the speed of sound and the wavelength and the frequency

                v = λ f

                f = v /λ

                f = 340 / 0.42

                f = 809.5 Hz

b) in this case the intensity is maximum

         Lam = Δr 2π / 2π

          Lam = 0.21 m

          fF= v / lam

          f = 340 / 0.21

         f = 1619 Hz

50 inches should be the resultant vector

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.