A particle executes simple harmonic motion represented by displacement function as x(t) = A sin( \omega t + \phi ) If the position and velocity of the particle at t = 0 s are 2 cm and 2 \omega cm s - 1 respectively, then its amplitude is x\sqrt 2 cm where the value of x is \underline\hspace2cm .

Simple Harmonic Motion question

2021 · 27 Jul · Shift 2 · Q66

2021 · 27 Jul · Shift 2 · Q66

View written solutionFree

Correct answer: 2

The SHM displacement is given by $x(t)=A\sin(\omega t+\phi).$
At $t=0$ : $x(0)=A\sin\phi=2\text{ cm}.$ So, $A\sin\phi=2. $
Velocity is the time derivative of displacement: $v(t)=\frac{dx}{dt}=A\omega \cos(\omega t+\phi).$
At $t=0$ : $v(0)=A\omega \cos\phi=2\omega\text{ cm s}^{-1}.$ Dividing by $\omega$ , $A\cos\phi=2.$
Now square and add the two equations: $A^2\sin^2\phi + A^2\cos^2\phi = 2^2+2^2.$ $A^2(\sin^2\phi+\cos^2\phi)=8.$ $A^2=8.$ $A=2\sqrt{2}\text{ cm}.$
The amplitude is given as $x\sqrt{2}$ cm. Comparing: $x\sqrt{2}=2\sqrt{2} \implies x=2.$

Therefore, the required integer is: $\boxed{2}$

2021 · 27 Jul · Shift 2 · Q66

2021 · 27 Jul · Shift 2 · Q66

View written solutionFree

Correct answer: 2

The SHM displacement is given by $x(t)=A\sin(\omega t+\phi).$
At $t=0$ : $x(0)=A\sin\phi=2\text{ cm}.$ So, $A\sin\phi=2. $
Velocity is the time derivative of displacement: $v(t)=\frac{dx}{dt}=A\omega \cos(\omega t+\phi).$
At $t=0$ : $v(0)=A\omega \cos\phi=2\omega\text{ cm s}^{-1}.$ Dividing by $\omega$ , $A\cos\phi=2.$
Now square and add the two equations: $A^2\sin^2\phi + A^2\cos^2\phi = 2^2+2^2.$ $A^2(\sin^2\phi+\cos^2\phi)=8.$ $A^2=8.$ $A=2\sqrt{2}\text{ cm}.$
The amplitude is given as $x\sqrt{2}$ cm. Comparing: $x\sqrt{2}=2\sqrt{2} \implies x=2.$

Therefore, the required integer is: $\boxed{2}$