Question

The maximum velocity of a harmonic oscillator is d and its maximum acceleration is $\beta$. Its time period will be

• A. $\frac{\pi \beta }{d}$
• B. $2\pi d\beta$
• C. $\frac{2\pi d}{\beta }$
• D. $\frac{2\pi \beta }{d}$

Answer 1

The correct option is C $\frac{2\pi d}{\beta }$

We know that $x=Asin\omega t$
Thus $v=dx/dt=A\omega cos\omega t$
and $a=dv/dt=-A{\omega }^{2}sin\omega t$
Max. v occurs when cosine is 1, which is $v_{max}=A\omega$
Max. a occurs when sine is 1 (or -1), which is $a_{max}=A{\omega }^2$ (magnitude)
Thus, $d=A\omega$, $\beta =A{\omega }^2$
Solve to get, $A{\omega }^2/A\omega =\beta /d$
Thus, $\omega =\beta /d$
or, $T=2\pi /\omega =2\pi d/\beta$