Question

A particle is subjected to SHM as given by equations $x_1=A_1\sin \omega t$ and $x_2=A_2\sin (\omega t+\pi /3).$The maximum acceleration and amplitude of the resultant motion
are $a_{max}$ and $A,$ respectively, then

• A. $a_{max}={\omega }^2\sqrt{A_1^2+A_2^2+A_1{A}_2}$
• B. $a_{max}={\omega }^2\sqrt{A_1{A}_2}$
• C. $A=A_1+A_2$
• D. $A=\sqrt{A_1^2+A_2^2+A_1{A}_2}$

Answer 1

The correct option is A $a_{max}={\omega }^2\sqrt{A_1^2+A_2^2+A_1{A}_2}$

The resultant of two motions is simple harmonic of same angular frequency $\omega$.
The amplitude of the resultant motion is
$A=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \frac{\pi }{3}}=\sqrt{A_1^2+A_2^2+A_1{A}_2}$
Maximum acceleration $={\omega }^{2}A={\omega }^2\sqrt{A_1^2+A_2^2+A_1{A}_2}$