Solution: Stroboscopic Effect on a Rotating Fan
Given:
- Frequency of fan rotation, $n_0 = 20 \text{ Hz}$.
- Flash rate range: $2 \le f_s \le 200$.
- Persistence of vision: $t_p \approx 1/16 \text{ s}$.
(a) Condition for the fan to appear stationary with 3 blades
For a 3-blade fan to appear stationary, the fan must rotate by an angle corresponding to its symmetry (or multiples thereof) between two consecutive flashes. The angle of symmetry is $\Delta \theta = \frac{2\pi}{3}$.
Let $T_s = 1/f_s$ be the time interval between flashes. The angle rotated by the fan in this time is:
$$ \theta_{rot} = \omega T_s = (2\pi n_0) \frac{1}{f_s} $$
For the blades to appear in indistinguishable positions, $\theta_{rot}$ must be an integer multiple of $120^\circ$ ($2\pi/3$ radians):
$$ \frac{2\pi n_0}{f_s} = k \frac{2\pi}{3} \quad \text{where } k \in \{1, 2, 3, \dots\} $$
$$ f_s = \frac{3n_0}{k} $$
Substituting $n_0 = 20$ Hz:
$$ f_s = \frac{60}{k} $$
We check for integer values of $f_s$ within the range $[2, 200]$:
For $k=1, f_s=60$; $k=2, f_s=30$; $k=3, f_s=20$; … up to $k=30, f_s=2$.
Flash rates: $\frac{60}{k}$ where $k \in \{1, 2, 3, 5, 6, 10, 15, 30\}$.
(b) Condition for the fan to appear stationary with 6 blades
To see 6 blades, the eyes must integrate images of the blades at their original positions AND at the intermediate positions (bisecting the $120^\circ$ angle). This effectively requires the fan to rotate by odd multiples of $60^\circ$ ($\pi/3$) between flashes.
Condition:
$$ \frac{2\pi n_0}{f_s} = (2k + 1) \frac{\pi}{3} \quad \text{where } k \in \{0, 1, 2, \dots\} $$
$$ \frac{2 n_0}{f_s} = \frac{2k + 1}{3} \implies f_s = \frac{6n_0}{2k + 1} $$
Substituting $n_0 = 20$ Hz:
$$ f_s = \frac{120}{2k + 1} $$
Valid integer values for range $[2, 200]$:
$k=1 \implies f_s = 40$ Hz
$k=2 \implies f_s = 24$ Hz
(Note: $k=0$ gives 120 Hz, but taking persistence of vision into consideration $k=1, 2$).
(c) Fan appears rotating in opposite direction with frequency $n = 4$ Hz
This is a stroboscopic aliasing effect. The fan appears to rotate backwards if the actual rotation falls slightly short of a full symmetry step.
Apparent frequency $n_{app} = f_s – n_{fan}$ (simplified). More formally, the angular shift per flash $\Delta \phi$ causing the apparent backward rotation $n$ is:
$$ \Delta \phi = 2\pi n T_s = \frac{2\pi n}{f_s} $$
The total rotation must be a symmetry step minus this small shift:
$$ \frac{2\pi n_0}{f_s} = k\frac{2\pi}{3} – \frac{2\pi n}{f_s} $$
$$ \frac{2\pi (n_0 + n)}{f_s} = \frac{2\pi k}{3} $$
$$ f_s = \frac{3(n_0 + n)}{k} $$
Substitute $n_0 = 20, n = 4$:
$$ f_s = \frac{3(24)}{k} = \frac{72}{k} $$
Valid integers for $k \in \{1, 2, 3, 4, 6, 8, 9, 12, 18, 36\}$.
The solutions should be different from part a so $k \in \{8, 9\}$.
Given:
- Frequency of fan rotation, $n_0 = 20 \text{ Hz}$.
- Flash rate range: $2 \le f_s \le 200$.
- Persistence of vision: $t_p \approx 1/16 \text{ s}$.
(a) Condition for the fan to appear stationary with 3 blades
For a 3-blade fan to appear stationary, the fan must rotate by an angle corresponding to its symmetry (or multiples thereof) between two consecutive flashes. The angle of symmetry is $\Delta \theta = \frac{2\pi}{3}$.
Let $T_s = 1/f_s$ be the time interval between flashes. The angle rotated by the fan in this time is:
$$ \theta_{rot} = \omega T_s = (2\pi n_0) \frac{1}{f_s} $$For the blades to appear in indistinguishable positions, $\theta_{rot}$ must be an integer multiple of $120^\circ$ ($2\pi/3$ radians):
$$ \frac{2\pi n_0}{f_s} = k \frac{2\pi}{3} \quad \text{where } k \in \{1, 2, 3, \dots\} $$ $$ f_s = \frac{3n_0}{k} $$Substituting $n_0 = 20$ Hz:
$$ f_s = \frac{60}{k} $$We check for integer values of $f_s$ within the range $[2, 200]$:
Flash rates: $\frac{60}{k}$ where $k \in \{1, 2, 3, 5, 6, 10, 15, 30\}$.
(b) Condition for the fan to appear stationary with 6 blades
To see 6 blades, the eyes must integrate images of the blades at their original positions AND at the intermediate positions (bisecting the $120^\circ$ angle). This effectively requires the fan to rotate by odd multiples of $60^\circ$ ($\pi/3$) between flashes.
Condition:
$$ \frac{2\pi n_0}{f_s} = (2k + 1) \frac{\pi}{3} \quad \text{where } k \in \{0, 1, 2, \dots\} $$ $$ \frac{2 n_0}{f_s} = \frac{2k + 1}{3} \implies f_s = \frac{6n_0}{2k + 1} $$Substituting $n_0 = 20$ Hz:
$$ f_s = \frac{120}{2k + 1} $$Valid integer values for range $[2, 200]$:
$k=2 \implies f_s = 24$ Hz
(Note: $k=0$ gives 120 Hz, but taking persistence of vision into consideration $k=1, 2$).
(c) Fan appears rotating in opposite direction with frequency $n = 4$ Hz
This is a stroboscopic aliasing effect. The fan appears to rotate backwards if the actual rotation falls slightly short of a full symmetry step.
Apparent frequency $n_{app} = f_s – n_{fan}$ (simplified). More formally, the angular shift per flash $\Delta \phi$ causing the apparent backward rotation $n$ is:
$$ \Delta \phi = 2\pi n T_s = \frac{2\pi n}{f_s} $$The total rotation must be a symmetry step minus this small shift:
$$ \frac{2\pi n_0}{f_s} = k\frac{2\pi}{3} – \frac{2\pi n}{f_s} $$ $$ \frac{2\pi (n_0 + n)}{f_s} = \frac{2\pi k}{3} $$ $$ f_s = \frac{3(n_0 + n)}{k} $$Substitute $n_0 = 20, n = 4$: