Question

A particle moves according to the equation x = Asin(\(\omega t\)). The potential energy is maximum at time \(t = \frac{T}{2\beta}\), where T is the time period of particle. Find the minimum value of \(\beta\) :

Hint:

Remember the standard SHM timeline: at \(t=0\), \(x=0\) (min PE); at \(t=T/4\), \(x=A\) (max PE); at \(t=T/2\), \(x=0\) (min PE); at \(t=3T/4\), \(x=-A\) (max PE).

Answer 1

Correct Answer: 2

To solve for the minimum value of β, we begin by considering the relationship between different energy forms in the given system. The equation of motion for the particle is \(x = A\sin(\omega t)\). The kinetic energy (KE) and potential energy (PE) of the particle are connected through this motion.

The potential energy is maximum when the displacement is maximum (i.e., \(x = A\)) because PE is proportional to \(x^2\). The potential energy, U, is given by:

\(U = \frac{1}{2}k x^2\), where \(k\) is the spring constant.

Given that the complete oscillation period \(T\) is the time period \(T = \frac{2\pi}{\omega}\), we know the condition for maximum potential energy is \(t = \frac{T}{4}\) but observed at \(t = \frac{T}{2\beta}\).

For PE to be maximum at \(t = \frac{T}{2\beta}\), \(t\) values of these points should align. Therefore:

\(\frac{T}{2\beta} = \left(n + \frac{1}{4}\right)T, \text{ for integer } n\)

Simplifying

\(\frac{1}{2\beta} = n + \frac{1}{4}\)

\(\Rightarrow 2\beta = \frac{1}{n + \frac{1}{4}}\)

\(\Rightarrow \beta = \frac{1}{2(n + \frac{1}{4})}\)

For the minimum value, let \(n = 0\),

\(\beta = \frac{1}{2 \times \frac{1}{4}} = 2\)

Thus, the minimum value of \(\beta\) is 2, confirming it is within the provided range which is [2, 2].