On the computer, we can only approximate the true Fourier transform because of the need for both time and frequency sampling. Because of the limited frequency resolution, the frequency of a cosine may not line up exactly at a frequency sample, but vou'll still get a peak around that frequency. Assuming that vou sample the signal with fs-8000 Hz, and take its Fourier transform using an FFT with 4096 points, what is the value (in Hz) of the FFT frequency bins closest to the two frequencies of the sinusoids (941 and 1336 Hz)

hello your question is incomplete below is the missing part

Consider the sum of sinusoids de (t) = sin(24(941)t) + sin(27(1336)t)

answer : Value of DFT frequency = 684 Hz

Explanation:

Given data:

sample signal ( Fs ) = 8000 Hz

Fourier transform using an FFT with 4096 points

Determine the value of DFT frequency bins closest to the two frequencies of the sinusoids ( 941 and 1336 Hz )    ( in HZ )

fs = 8000 Hz

4096 parts are divided with a gap of Fs/N

 attached below is the detailed solution

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft