A particle is oscillating in a stright line about a centre of force `O`, towards which when at a distance `x` the force is `mn^(2)x` where m is the mass, n a constant. The amplitude is `a = 15 cm`. When a distance `(asqrt(3))/(2)` from O, find the new amplitude.

Answer

To solve the problem step by step, we will follow the principles of simple harmonic motion (SHM) and the relationship between force, displacement, and amplitude.

### Step-by-Step Solution:

1. **Understand the Force Equation**:
The force acting on the particle is given by:
\[ F = mn^2 x \]
This indicates that the force is proportional to the displacement \( x \) from the center of force \( O \).

2. **Relate Force to SHM**:
In SHM, the force can also be expressed as:
\[ F = -m \omega^2 x \]
By comparing the two equations, we can equate:
\[ mn^2 x = -m \omega^2 x \]
This implies:
\[ \omega^2 = n^2 \]
Therefore, the angular frequency \( \omega \) is:
\[ \omega = n \]

3. **Velocity in SHM**:
The velocity \( v \) of a particle in SHM is given by:
\[ v = \omega \sqrt{a^2 - x^2} \]
where \( a \) is the amplitude and \( x \) is the displacement from the center.

4. **Calculate Velocity at Maximum Amplitude**:
At maximum amplitude \( x = 0 \):
\[ v = n \sqrt{a^2 - 0^2} = n a \]

5. **Calculate Velocity at Given Displacement**:
When the particle is at a distance \( x = \frac{a \sqrt{3}}{2} \):
\[ v = n \sqrt{a^2 - \left(\frac{a \sqrt{3}}{2}\right)^2} \]
Simplifying the expression:
\[ v = n \sqrt{a^2 - \frac{3a^2}{4}} = n \sqrt{\frac{a^2}{4}} = n \cdot \frac{a}{2} \]

6. **Relate Velocities**:
The new velocity \( v' \) at \( x = \frac{a \sqrt{3}}{2} \) is:
\[ v' = n \sqrt{A'^2 - x^2} \]
where \( A' \) is the new amplitude. Setting the two expressions for velocity equal gives:
\[ n \cdot \frac{a}{2} = n \sqrt{A'^2 - \left(\frac{a \sqrt{3}}{2}\right)^2} \]

7. **Cancel \( n \) and Square Both Sides**:
Cancel \( n \) from both sides:
\[ \frac{a}{2} = \sqrt{A'^2 - \frac{3a^2}{4}} \]
Squaring both sides:
\[ \left(\frac{a}{2}\right)^2 = A'^2 - \frac{3a^2}{4} \]
This simplifies to:
\[ \frac{a^2}{4} = A'^2 - \frac{3a^2}{4} \]

8. **Solve for the New Amplitude**:
Rearranging gives:
\[ A'^2 = \frac{a^2}{4} + \frac{3a^2}{4} = \frac{4a^2}{4} = a^2 \]
Therefore:
\[ A' = \sqrt{3} a \]

9. **Substituting the Initial Amplitude**:
Given the initial amplitude \( a = 15 \, \text{cm} \):
\[ A' = \sqrt{3} \cdot 15 \, \text{cm} = 15\sqrt{3} \, \text{cm} \]

### Final Answer:
The new amplitude is:
\[ A' = 15\sqrt{3} \, \text{cm} \]