Two pianos each sound the same note simultaneously, but they are both out of tune. On a day when the speed of sound is 343 m/s, piano A produces a wavelength of 0.762 m, while piano B produces a wavelength of 0.781 m. How much time separates successive beats?

time period = 0.09 s

Explanation:

given data

speed of sound v = 343 m/s

wavelength λ1 = 0.762 m

wavelength λ2 = 0.781  m

solution

we know that beat frequency is the difference between the frequency of the individual wave so

frequency f1 =   ..........................1

frequency f1 = = 450.13

and

frequency f2 =     = 439.18

so beat frequency = f1 - f2

beat frequency = 450.13 - 439.18

beat frequency = 10.95 Hz

and

time period between successive beats is

time period =  

time period =

time period = 0.09 s

see earlier answer

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .