A particle of mass 1 ~kg is subjected to a force which depends on the position as \vecF=-k(x \hat\imath+y \hat\jmath) kg\, m s^-2 with k=1 ~kg ~s^-2 . At time t=0 , the particle's position \vecr=(\frac1\sqrt2 \hat\imath+\sqrt2 \hat\jmath) m and its velocity \vecv=(-\sqrt2 \hat\imath+\sqrt2 \hat\jmath+\frac2\pi \hatk) m s^-1 . Let v_x and v_y denote the x and the y components of the particle's velocity, respectively. Ignore gravity. When z=0.5 ~m , the value of (x v_y-y v_x) is \underline\hspace2cm m^2 s^-1 .

Simple Harmonic Motion PYQ — JEE Advanced 2022 · Shift 2 · Q37 with Solution

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Correct answer: 3

Given force and equations of motion

The force is

\vec F=-k(x\hat i+y\hat j), \qquad k=1

and mass is $m=1\,\text{kg}$ .

So acceleration is

\vec a=\frac{\vec F}{m}=-(x\hat i+y\hat j)

Thus,

\ddot x=-x, \qquad \ddot y=-y, \qquad \ddot z=0

Hence $x$ and $y$ execute SHM with angular frequency

\omega=1\,\text{rad/s}

while motion along $z$ is uniform.

Initial conditions

At $t=0$ :

x_0=\frac{1}{\sqrt2}, \qquad y_0=\sqrt2

v_{x0}=-\sqrt2, \qquad v_{y0}=\sqrt2, \qquad v_{z0}=\frac{2}{\pi}

Also, from the given position vector, initial $z$ -coordinate is

z_0=0

Find the time when $z=0.5$ m

Since $\ddot z=0$ , we have

z=z_0+v_{z0}t=\frac{2}{\pi}t

Set $z=0.5$ :

\frac{2}{\pi}t=\frac12

t=\frac{\pi}{4}

Useful conserved quantity in 2D SHM

We need

xv_y-yv_x

Notice this is the $z$ -component of angular momentum per unit mass:

L_z/m = x\dot y-y\dot x

Differentiate it:

\frac{d}{dt}(x\dot y-y\dot x)=\dot x\dot y+x\ddot y-\dot y\dot x-y\ddot x

= x(-y)-y(-x)= -xy+xy=0

So,

xv_y-yv_x = \text{constant}

Therefore its value at any time equals its value at $t=0$ .

Evaluate at $t=0$

xv_y-yv_x=\left(\frac{1}{\sqrt2}\right)(\sqrt2)-\left(\sqrt2\right)(-\sqrt2)

=1-(-2)=3

Thus when $z=0.5\,\text{m}$ ,

xv_y-yv_x=3\,\text{m}^2\text{s}^{-1}

Comparison with stored answer

Derived answer = $3$

Stored correct answer = $3$

They match.

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View written solutionFree

Correct answer: 3

Given force and equations of motion

The force is

\vec F=-k(x\hat i+y\hat j), \qquad k=1

and mass is $m=1\,\text{kg}$ .

So acceleration is

\vec a=\frac{\vec F}{m}=-(x\hat i+y\hat j)

Thus,

\ddot x=-x, \qquad \ddot y=-y, \qquad \ddot z=0

Hence $x$ and $y$ execute SHM with angular frequency

\omega=1\,\text{rad/s}

while motion along $z$ is uniform.

Initial conditions

At $t=0$ :

x_0=\frac{1}{\sqrt2}, \qquad y_0=\sqrt2

v_{x0}=-\sqrt2, \qquad v_{y0}=\sqrt2, \qquad v_{z0}=\frac{2}{\pi}

Also, from the given position vector, initial $z$ -coordinate is

z_0=0

Find the time when $z=0.5$ m

Since $\ddot z=0$ , we have

z=z_0+v_{z0}t=\frac{2}{\pi}t

Set $z=0.5$ :

\frac{2}{\pi}t=\frac12

t=\frac{\pi}{4}

Useful conserved quantity in 2D SHM

We need

xv_y-yv_x

Notice this is the $z$ -component of angular momentum per unit mass:

L_z/m = x\dot y-y\dot x

Differentiate it:

\frac{d}{dt}(x\dot y-y\dot x)=\dot x\dot y+x\ddot y-\dot y\dot x-y\ddot x

= x(-y)-y(-x)= -xy+xy=0

So,

xv_y-yv_x = \text{constant}

Therefore its value at any time equals its value at $t=0$ .

Evaluate at $t=0$

xv_y-yv_x=\left(\frac{1}{\sqrt2}\right)(\sqrt2)-\left(\sqrt2\right)(-\sqrt2)

=1-(-2)=3

Thus when $z=0.5\,\text{m}$ ,

xv_y-yv_x=3\,\text{m}^2\text{s}^{-1}

Comparison with stored answer

Derived answer = $3$

Stored correct answer = $3$

They match.

Simple Harmonic Motion question

Simple Harmonic Motion question

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Simple Harmonic Motion question

Simple Harmonic Motion question

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