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

Explanation

To solve the problem step by step, we need to determine the maximum frequency of the simple harmonic motion (SHM) of the surface such that a mass placed on it remains in continuous contact.

### Step 1: Understand the relationship between forces
When the surface moves in SHM, the mass will remain in contact with the surface as long as the upward acceleration of the surface is greater than or equal to the acceleration due to gravity. The forces acting on the mass are:

- Weight of the mass (downward): \( mg \)
- Normal force (upward): \( N \)

For the mass to remain in contact with the surface, the normal force must be at least zero, which means the upward acceleration must be equal to or greater than \( g \).

### Step 2: Write the equation for SHM
The acceleration \( a \) of the surface in SHM can be expressed as:
\[ a = -\omega^2 x \]
where \( \omega \) is the angular frequency and \( x \) is the displacement from the mean position. The maximum acceleration occurs at maximum displacement (amplitude \( A \)):
\[ a_{\text{max}} = \omega^2 A \]

### Step 3: Set up the inequality
For the mass to remain in contact, we need:
\[ \omega^2 A \geq g \]
This means:
\[ \omega^2 \geq \frac{g}{A} \]

### Step 4: Solve for \( \omega \)
Taking the square root of both sides gives:
\[ \omega \geq \sqrt{\frac{g}{A}} \]

### Step 5: Substitute values
Given:
- \( g = 9.8 \, \text{m/s}^2 \)
- \( A = 1 \, \text{cm} = 0.01 \, \text{m} \)

Substituting these values into the equation:
\[ \omega \geq \sqrt{\frac{9.8}{0.01}} \]
\[ \omega \geq \sqrt{980} \]

### Step 6: Calculate \( \omega \)
Calculating \( \sqrt{980} \):
\[ \sqrt{980} \approx 31.3 \, \text{rad/s} \]

### Step 7: Relate \( \omega \) to frequency \( f \)
The relationship between angular frequency \( \omega \) and frequency \( f \) is given by:
\[ \omega = 2\pi f \]
Thus,
\[ f = \frac{\omega}{2\pi} \]

### Step 8: Substitute \( \omega \) into the frequency equation
Substituting the value of \( \omega \):
\[ f \geq \frac{31.3}{2\pi} \]
\[ f \geq \frac{31.3}{6.283} \]
\[ f \geq 4.98 \, \text{Hz} \]

### Step 9: Round to the nearest whole number
The maximum frequency \( f \) that allows the mass to remain in contact is approximately \( 5 \, \text{Hz} \).

### Final Answer
The maximum frequency of SHM is \( 5 \, \text{Hz} \).

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