A small source of sound executes simple harmonic motion with an...

The maximum velocity of the source executing simple harmonic motion is v_\max = A\omega = A\left(\dfrac{2\pi}{T}\right).

The maximum frequency observed by the detector occurs when the source moves towards it with maximum velocity:

f_\max = f_0 \left(\dfrac{v}{v - v_\max}\right)

The minimum frequency observed occurs when the source moves away from it with maximum velocity:

f_\min = f_0\left(\dfrac{v}{v + v_\max}\right)

The width of the frequency band is given by \Delta f = f_\max - f_\min:

\Delta f = f_0 v \left(\dfrac{1}{v - v_\max} - \dfrac{1}{v + v_\max} \right) =f_0 v \left( \dfrac{2v_\max}{v^2 - v_\max^2} \right)

Since v_\max is much smaller than $v$, we can approximate v^2 - v_\max^2 \approx v^2, which gives:

\Delta f \approx f_0 \left(\dfrac{2v_\max}{v} \right)

Substituting the given values $\Delta f = 8 \text{ Hz}$, $f_0 = 800 \text{ Hz}$, and $v = 340 \text{ m s}^{-1}$:

8 = 800 \left( \dfrac{2v_\max}{340} \right)

1 = 100 \left( \dfrac{2v_\max}{340} \right) \Rightarrow v_\max = \dfrac{340}{200} = 1.7 \text{ m s}^{-1}

Now, using the relation for maximum velocity in SHM with $A = 17 \text{ cm} = 0.17 \text{ m}$:

v_\max = A \left(\dfrac{2\pi}{T} \right)

$1.7 = 0.17 \times$\dfrac{2\pi}{T}

$10 =$\dfrac{2\pi}{T} \Rightarrow$T =$\dfrac{\pi}{5} \text{ s}

$T \approx 0.628 \text{ s} \approx 0.63 \text{ s}$