Question

A particle executes simple harmonic motion represented by displacement function as x(t) = A sin($\omega$t+$\phi$). If the position and velocity of the particle at t = 0 s are 2 cm and 2$\omega$ cm s$^{-1}$ respectively, then its amplitude is $x\sqrt{2}$ cm where the value of x is ________.

Hint:

A useful relation in SHM connects amplitude, position, and velocity: $A^2 = x^2 + (v/\omega)^2$. Using this formula with the initial conditions ($x_0 = 2$, $v_0 = 2\omega$) gives the answer directly: $A^2 = 2^2 + (2\omega/\omega)^2 = 4 + 4 = 8$.

Answer 1

Correct Answer: 2

The displacement of the particle in SHM is given by $x(t) = A \sin(\omega t + \phi)$.
The velocity of the particle is the time derivative of the displacement:
$v(t) = \frac{dx}{dt} = A\omega \cos(\omega t + \phi)$.
We are given the initial conditions at $t=0$:
Position, $x(0) = 2$ cm.
Velocity, $v(0) = 2\omega$ cm/s.
Substitute $t=0$ into the equations for x(t) and v(t):
$x(0) = A \sin(0 + \phi) = A \sin\phi$.
$v(0) = A\omega \cos(0 + \phi) = A\omega \cos\phi$.
Using the given values, we have two equations:
$A \sin\phi = 2$. (Equation 1)
$A\omega \cos\phi = 2\omega \implies A \cos\phi = 2$. (Equation 2)
To find the amplitude A, we can square and add Equation 1 and Equation 2.
$(A \sin\phi)^2 + (A \cos\phi)^2 = 2^2 + 2^2$.
$A^2 \sin^2\phi + A^2 \cos^2\phi = 4 + 4$.
$A^2 (\sin^2\phi + \cos^2\phi) = 8$.
Since $\sin^2\phi + \cos^2\phi = 1$:
$A^2 = 8$.
$A = \sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$ cm.
The problem states that the amplitude is given in the form $x\sqrt{2}$ cm.
Comparing our result $A = 2\sqrt{2}$ cm with the given form $A = x\sqrt{2}$ cm, we can see that:
$x = 2$.