Question

Which of the following quantities is doubled on doubling the amplitude of a harmonic oscillator

• A. $E_t$
• B. $E_k$
• C. $v_{max}$
• D. U

Answer 1

Answer: (C)

$v_{max}$

The total mechanical energy of S.H.O is given E $=$ P.E. + K.E.

$=$ $\frac{1}{2}m{\omega }^2{x}^2$ + $\frac{1}{2}m{\omega }^2(A^2-x^2)$

$=$ $\frac{1}{2}m{\omega }^2{A}^2$

$P.E.$ $=$ $\frac{1}{2}m{\omega }^2{x}^2$ $=$ $2{\pi }^{2}m{v}^2{x}^2$

Where $v$ $=$ frequency of oscillation

P.E. can be written as

$P.E.$ $=$ $\frac{1}{2}m{\omega }^2{A}^2{\sin }^2\omega t$

(Where $x$ $=$ $A\sin \omega t$)

The reference P.E. of the S.H.M. can be taken as zero at the mean position.

P.E. of the S.H.M. is maximum at the extreme position.

$x=Asin(wt+\varphi )$$v=-AwCos(wt+\varphi )$

$\alpha =-A{w}^{2}Sin(wt+\varphi )=-w^{2}x$

$v=w\sqrt{A^2-x^2}$

$v_{max}=wA$,

$a_{max}=w^{2}A$

The total mechanical energy of the system and U(P.E) of system will become four times when the amplitude is doubled, as seen from above equations 1 and 2. Also, K.E is not doubled when A is doubled, from formula 1. By observing Formula 3 and options, we see that maximum velocity doubles up when we double the amplitude. Hence, option C is correct.