A sinusoidal transverse wave travels along a long, stretched string. The amplitude of this wave is 0.0957 m, its frequency is 3.75 Hz, and its wavelength is 1.97 m. What is the shortest transverse distance between a maximum and a minimum of the wave?

Shortest distance = 0.1914 m

Explanation:

Given that,

Amplitude of the wave is 0.0957 m

Frequency of the wave is 3.75 Hz

Wavelength of the wave is 1.97 m

We need to find the shortest transverse distance between a maximum and a minimum of the wave.

The distance between maximum point in positive axis and the baseline is equal to amplitude.

Shortest distance = 2 A

D = 2 × 0.0957

D = 0.1914 m

So, the shortest transverse distance between a maximum and a minimum of the wave is 0.1914 m.

the inner core.

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klicker

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 .