Question

please solve this question with explanation​

Answer 1

Let the frequencies of the tuning forks are in AP with common difference -x (since they're arranged in decreasing order of frequency).

Let the frequency of first tuning fork be,

a(1) = n

So the frequency of p'th tuning fork is,

a(p) = n - (p - 1)x

Thus the frequency of the last (33rd) tuning fork is,

a(33) = n - 32x → (1)

We have the frequency of the 10th tuning fork,

a(10) = 440

n - 9x = 440

n = 9x + 440 → (2)

Given that first fork is octave of the last. It means the frequency of the first fork is double that of the last fork. Hence,

a(1) = 2 · a(33)

n = 2(n - 32x)

n = 2n - 64x

n = 64x

9x + 440 = 64x [From (2)]

x = 8 Hz

Thus the frequency of the first fork is, from (2),

a(1) = 9 × 8 + 440

a(1) = 512 Hz

And that of the last fork is, from (1),

a(33) = 512 - 32 × 8

a(33) = 256 Hz

Answer 2

$\mathfrak{ \huge{ \red{ \underline{ \underline{Answer}}}}}$

• Given :

=> A set of 33 tuning fork is arranged in series with decreasing order of frequency.

=> Each fork produces x beats per second with previous one.

=> Freq. of first fork is double than freq. of last fork.

=> Freq. of 10th fork is 440 Hz.

• To Find :

=> Freq. of first and last fork

• Calculation :

=> let total number of forks = n

=> freq. of first fork = f1

=> freq. of 10th fork f10 = 440 Hz = f1 - 9x

=> freq. of 33th fork f33 = f1 - 32x

=> f33 = f1/2 therefore f1 = 64x

=> 440 = 64x - 9x = 55x

=> x = no. of beats = 440/55 = 8

=> Freq. of first fork = 64x = 512Hz

=> Freq. of last fork = 512 - 32(8) = 256Hz

$\boxed{ \boxed{ \blue{ \bold{ \sf{ \huge{f(1) = 512 \: Hz \: and \: f(33) = 256 \: Hz}}}}}}$