A simple harmonic oscillator consists of a mass attached to a spring. The mass moves back and forth due to the restoring force exerted by the spring. The parameters given for this system are the mass \(m\) kg and the spring constant \(k\) Nm-1.
The equation for a simple harmonic oscillator is given by:
\(F = -kx = ma\)
The acceleration \(a\) can be linked to velocity \(v\) and position \(x\) using the energy conservation principle:
\(\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \text{constant}\)
The equation above signifies that the sum of kinetic and potential energy remains constant, representing an ellipse in the \(x-v\) phase space.
For the graph of \(v\) as a function of \(x\) to be a circle, the equation should be of the form where the coefficient of \(x^2\) and \(v^2\) are equal. Thus:
\(\frac{1}{2}mv^2 + \frac{1}{2}kx^2 = E \quad \Rightarrow \quad mv^2 = kx^2\)
For the coefficients of \(v^2\) and \(x^2\) to be equal, we require:
\(m = k\)
Therefore, for the graph to be a circle, the relationship \(k = m\) must hold true.
Hence, the correct option is:
$k=m$
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A particle is subjected to two simple harmonic motions as :
$x_1 = \sqrt{7} \sin 5t \text{ cm}$ and $x_2 = 2\sqrt{7} \sin \left(5t + \frac{\pi}{3}\right) \text{ cm}$
where x is displacement and t is time in seconds. The maximum acceleration of the particle is $x \times 10^{-2} \text{ ms}^{-2}$. The value of x is :- The equation of a transverse wave is $y = y_0 \sin 2\pi(ft - \frac{x}{\lambda})$. If the maximum particle velocity be four times that of wave velocity then
