A large horizontal surface moves up and down in SHM with an amplitude of 1 cm . If a mass of 10 kg (which is placed on the surface) is to remain continually in contact with it, the maximum frequency of S.H.M. will be

Options

A. 0.5 Hz
B. 1.5 Hz
C. 5 Hz
D. 10 Hz

Answer

To find the maximum frequency of simple harmonic motion (SHM) for a mass to remain in contact with a surface moving up and down, we can follow these steps:

### Step 1: Understand the conditions for contact
For the mass to remain in contact with the surface, the upward acceleration of the surface must be greater than or equal to the acceleration due to gravity. This ensures that the mass does not lose contact with the surface.

### Step 2: Write the expression for acceleration in SHM
The acceleration \( a \) of an object in SHM is given by the formula:
\[ a = -\omega^2 x \]
where:
- \( \omega \) is the angular frequency in radians per second,
- \( x \) is the displacement from the mean position.

### Step 3: Determine the maximum displacement
In this case, the maximum displacement \( x \) is equal to the amplitude \( A \). Given that the amplitude \( A = 1 \, \text{cm} = 0.01 \, \text{m} \).

### Step 4: Set up the inequality for contact
The upward acceleration must be at least equal to the acceleration due to gravity \( g \):
\[ \omega^2 A \geq g \]
Substituting the values:
\[ \omega^2 (0.01) \geq 9.81 \]

### Step 5: Solve for \( \omega \)
Rearranging the inequality gives:
\[ \omega^2 \geq \frac{9.81}{0.01} \]
\[ \omega^2 \geq 981 \]
Taking the square root:
\[ \omega \geq \sqrt{981} \approx 31.34 \, \text{rad/s} \]

### Step 6: Convert angular frequency to frequency
The frequency \( f \) is related to angular frequency \( \omega \) by the formula:
\[ f = \frac{\omega}{2\pi} \]
Substituting the value of \( \omega \):
\[ f \geq \frac{31.34}{2\pi} \approx \frac{31.34}{6.2832} \approx 5 \, \text{Hz} \]

### Conclusion
The maximum frequency of SHM for the mass to remain continually in contact with the surface is approximately **5 Hz**.

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