The differential equation of the simple harmo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the equation of simple harmonic motion: $x = A \cos (nt + \alpha)$ ... $(i)$ Differentiating with respect to $t$: $\frac{dx}{dt} = -A n \sin

Vedclass · Accepted Answer

$\frac{d^2 x}{d t^2}+n^2 x=0$

Vedclass · Answer

(B) Given the equation of simple harmonic motion: $x = A \cos (nt + \alpha)$ ... $(i)$ Differentiating with respect to $t$: $\frac{dx}{dt} = -A n \sin (nt + \alpha)$ ... (ii) Differentiating again with respect to $t$: $\frac{d^2x}{dt^2} = -An \frac{d}{dt} \sin (nt + \alpha)$ $\frac{d^2x}{dt^2} = -An^2 \cos (nt + \alpha)$ Substituting $x = A \cos (nt + \alpha)$ into the equation: $\frac{d^2x}{dt^2} = -n^2 x$ Therefore, the differential equation is: $\frac{d^2x}{dt^2} + n^2 x = 0$

The differential equation of the simple harmonic motion given by $x=A \cos (n t+\alpha)$ is

Similar Questions

The differential equation of $3y = \sqrt[3]{x} + c$ is

Verify that the given function $y=\sqrt{1+x^{2}}$ is a solution of the differential equation $y^{\prime}=\frac{xy}{1+x^{2}}$.

The differential equation formed by eliminating arbitrary constants $A$ and $B$ from the equation $y = A \cos 3x + B \sin 3x$ is

The differential equation for which $y = ax^2 + bx + c$ is the general solution is:

Find the differential equation for the family of curves $y^{2}=a(b^{2}-x^{2})$ by eliminating the arbitrary constants $a$ and $b$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module