A light pointer fixed to one prong of a tuning fork touches gnetly a smoked vertical plate. The fork is set vibrating and the plate is allowed to fall freely. 8 complete oscilllations are counted when the plate falls through 10cm.What is the frequency of the tuning fork?

Options

A. 112 Hz
B. 56 Hz
C. 8/7 Hz
D. 7/8 Hz

Answer

To solve the problem step-by-step, we will use the principles of kinematics and the definition of frequency.

### Step 1: Understand the problem
A tuning fork is vibrating, and its pointer touches a smoked vertical plate. The plate falls freely, and during its fall, 8 complete oscillations are counted when the plate falls through a distance of 10 cm.

### Step 2: Identify the known values
- Distance fallen (s) = 10 cm = 0.1 m (convert to meters)
- Number of oscillations (n) = 8
- Initial velocity (u) = 0 (the plate starts from rest)
- Acceleration (a) = g = 9.81 m/s² (acceleration due to gravity)

### Step 3: Use the kinematic equation
We can use the kinematic equation for displacement:
\[ s = ut + \frac{1}{2} a t^2 \]

Substituting the known values:
\[ -0.1 = 0 \cdot t + \frac{1}{2} (-9.81) t^2 \]
This simplifies to:
\[ -0.1 = -4.905 t^2 \]

### Step 4: Solve for time (t)
Rearranging the equation gives:
\[ t^2 = \frac{0.1}{4.905} \]
Calculating this:
\[ t^2 = 0.0204 \]
Taking the square root:
\[ t = \sqrt{0.0204} \approx 0.142 \text{ seconds} \]

### Step 5: Calculate the frequency
Frequency (f) is defined as the number of oscillations per unit time. We have 8 oscillations in approximately 0.142 seconds:
\[ f = \frac{n}{t} = \frac{8}{0.142} \]

Calculating this gives:
\[ f \approx 56.34 \text{ Hz} \]

### Step 6: Round the frequency
Since frequency is typically rounded to the nearest whole number:
\[ f \approx 56 \text{ Hz} \]

### Conclusion
The frequency of the tuning fork is approximately **56 Hz**.

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