Deriving the Differential Equation from x = A cos(mt - α)

We are given the relation for \(x\) as a function of time \(t\):

\(x = A \cos(mt - \alpha)\)

Here, \(A\), \(m\), and \(\alpha\) are constants. We need to find the differential equation that this relation satisfies. This means we need to find a relationship between \(x\) and its derivatives with respect to \(t\).

Let's find the first derivative of \(x\) with respect to \(t\), \(\frac{dx}{dt}\). We use the chain rule:

\(\frac{dx}{dt} = \frac{d}{dt}(A \cos(mt - \alpha))\)

\(\frac{dx}{dt} = A \cdot (-\sin(mt - \alpha)) \cdot \frac{d}{dt}(mt - \alpha)\)

\(\frac{dx}{dt} = -A \sin(mt - \alpha) \cdot m\)

\(\frac{dx}{dt} = -Am \sin(mt - \alpha)\)

Now, let's find the second derivative of \(x\) with respect to \(t\), \(\frac{d^2x}{dt^2}\). We differentiate \(\frac{dx}{dt}\) with respect to \(t\):

\(\frac{d^2x}{dt^2} = \frac{d}{dt}(-Am \sin(mt - \alpha))\)

\(\frac{d^2x}{dt^2} = -Am \cdot (\cos(mt - \alpha)) \cdot \frac{d}{dt}(mt - \alpha)\)

\(\frac{d^2x}{dt^2} = -Am \cos(mt - \alpha) \cdot m\)

\(\frac{d^2x}{dt^2} = -Am^2 \cos(mt - \alpha)\)

We have the second derivative as \(\frac{d^2x}{dt^2} = -Am^2 \cos(mt - \alpha)\). Notice that the original expression for \(x\) was \(x = A \cos(mt - \alpha)\). We can substitute this back into the equation for the second derivative:

\(\frac{d^2x}{dt^2} = -m^2 (A \cos(mt - \alpha))\)

\(\frac{d^2x}{dt^2} = -m^2 x\)

This is the differential equation that satisfies the given relation \(x = A \cos(mt - \alpha)\). This equation is a standard form for Simple Harmonic Motion (SHM), where \(m^2\) represents the square of the angular frequency.

Now let's compare this derived differential equation with the given options:

• Option 1: \(\frac{d^2x}{dt^2} = -\alpha^2x\) - Does not match \( -m^2 x \).
• Option 2: \(\frac{d^2x}{dt^2} = -m^2x\) - Matches our derived equation.
• Option 3: \(\frac{d^2x}{dt^2} = m^2x\) - Does not match \(-m^2 x \).
• Option 4: \(\frac{dx}{dt} = 1 - x^2\) - This is a first-order differential equation and does not match the form or the result we obtained.

Therefore, the differential equation satisfying the relation \(x = A \cos(mt - \alpha)\) is \(\frac{d^2x}{dt^2} = -m^2x\).