A point mass is subjected to two simultaneous sinusoidal displacement in `x` -direction, `x_(1)(t)=A sin omega t` and `x_(2)(t)=A sin (omega t+(pi)/(2))`. Adding a third sinusoidal displacement `x_(3)t=B sin (omega t+phi)` brings the mass to a complete rest. The value of `phi` is `n (pi)/(4)`. Then `n` is.

Answer

To solve the problem, we need to analyze the given sinusoidal displacements and find the value of \( n \) such that the third displacement \( x_3 \) brings the mass to a complete rest.

### Step 1: Understand the given displacements
We have two displacements:
1. \( x_1(t) = A \sin(\omega t) \)
2. \( x_2(t) = A \sin\left(\omega t + \frac{\pi}{2}\right) \)

The second displacement can be rewritten using the sine function:
\[ x_2(t) = A \cos(\omega t) \]
This is because \( \sin\left(\theta + \frac{\pi}{2}\right) = \cos(\theta) \).

### Step 2: Resultant of the first two displacements
The resultant displacement \( x_r(t) \) of the two displacements can be found using vector addition:
\[ x_r(t) = x_1(t) + x_2(t) = A \sin(\omega t) + A \cos(\omega t) \]

### Step 3: Express the resultant in a single sinusoidal form
To combine \( A \sin(\omega t) \) and \( A \cos(\omega t) \), we can use the following formula:
\[ R = \sqrt{A^2 + A^2} = \sqrt{2A^2} = A\sqrt{2} \]
The angle \( \theta \) can be found using:
\[ \tan(\theta) = \frac{A}{A} = 1 \]
Thus, \( \theta = \frac{\pi}{4} \).

Therefore, we can express the resultant as:
\[ x_r(t) = A\sqrt{2} \sin\left(\omega t + \frac{\pi}{4}\right) \]

### Step 4: Introduce the third displacement
We introduce the third displacement:
\[ x_3(t) = B \sin\left(\omega t + \phi\right) \]
We want the total displacement to be zero for the mass to come to rest:
\[ x_r(t) + x_3(t) = 0 \]

### Step 5: Condition for complete rest
For the mass to come to rest, the amplitude of the third displacement must equal the amplitude of the resultant displacement but in the opposite direction:
\[ B \sin\left(\omega t + \phi\right) = -A\sqrt{2} \sin\left(\omega t + \frac{\pi}{4}\right) \]

### Step 6: Determine the phase difference
To achieve this, we need:
\[ \phi = \frac{\pi}{4} + \pi = \frac{5\pi}{4} \]

### Step 7: Find the value of \( n \)
We are given that \( \phi = n \frac{\pi}{4} \). Setting this equal to our found value:
\[ n \frac{\pi}{4} = \frac{5\pi}{4} \]

Dividing both sides by \( \frac{\pi}{4} \):
\[ n = 5 \]

### Final Answer
Thus, the value of \( n \) is:
\[ \boxed{5} \]