QBManager

Oscillations

139997 For particle $P$ revolving round the centre $O$ with radius of circular path $r$ and angular velocity $ω$ as shown in below figure, the projection of OP on the $x$ -axis at time $t$ is

1

x (t) = r \cos (ω t + \frac{π}{6})

2

x (t) = r \cos (ω t)

3

x (t) = r \sin (ω t + \frac{π}{6})

4

x (t) = r \cos (ω t - \frac{π}{6} ω)

Explanation:

A We know that in polar coordinate the horizontal component
$x = r \cos θ$
But given here $θ = (ω t + ϕ) = 30^{\circ}$
Therefore, putting the value of $θ$ in above equation

$x (t) = r \cos (ω t + 30^{\circ})$
$x (t) = r \cos (ω t + π / 6)$
Hence, projection of OP on $x$ -axis $= r \cos (ω t + π / 6)$

JEE Main-08.04.2023

Oscillations

139999 In a linear simple harmonic motion (SHM)

1 (A), (B) and (C) only

2 (C) and (D) only

3 (A), (B) and (D) only

4 (A), (C) and (D) only

Explanation:

A
We know that, the restoring force in simple harmonic motion
$∵ F = - kx$
- And acceleration in SHM $a = ω^{2} x$
- Velocity is maximum at mean position
- Acceleration is maximum at extreme points

JEE Main-15.04.2023

Oscillations

140000 The displacement of a particle executing SHM is given by $x = 3 \sin [2 π t + \frac{π}{4}]$ where ' $x$ ' is in meters and ' $t$ ' is in seconds. The amplitude and maximum speed of the particle is

1

3 m, 6 π {ms}^{- 1}

2

3 m, 8 π {ms}^{- 1}

3

3 m, 2 π {ms}^{- 1}

4

3 m, 4 π {ms}^{- 1}

Explanation:

A Given,
$x = 3 \sin [2 π t + \frac{π}{4}]$
As we know that the standard equation of SHM is -
$x = A \sin [ω t + ϕ]$
On comparing equation (i) and (ii), we get -
$A = 3 m, ω = 2 π, ϕ = \frac{π}{4}$
$∵ v_{max} = A ω$
$v_{max} = 3 \times 2 π$
$v_{max} = 6 π {ms}^{- 1}$
The amplitude and maximum speed of particle is $3 m$ , $6 π {ms}^{- 1}$ respectively.

Karnataka CET-2022

Oscillations

140001 A mass $m$ is performing linear simple harmonic motion, then which of the following graph represents correctly the variation of acceleration 'a' corresponding to linear velocity 'v' ?

1

2

3

4

Explanation:

D In simple harmonic motion,
$y = r \sin ω t$
$and v = r ω \cos ω t \Rightarrow \cos ω t = \frac{v}{r ω}$
$a = \frac{d v}{a t} = \frac{d}{d t} (r ω \cos ω t)$
$a = - ω^{2} r \sin ω t$
$\sin ω t = - \frac{a}{ω^{2} r}$
Hence,
$\sin^{2} ω t + \cos^{2} ω t = {(- \frac{a}{ω^{2} r})}^{2} + {(\frac{v}{r ω})}^{2}$
$1 = \frac{a^{2}}{ω^{4} r^{2}} + \frac{v^{2}}{r^{2} ω^{2}} [∵ \sin^{2} θ + \cos^{2} θ = 1]$
$- \frac{a^{2}}{ω^{4} r^{2}} = \frac{v^{2}}{r^{2} ω^{2}} - 1$
$- \frac{a^{2}}{ω^{4} r^{2}} = \frac{v^{2} - r^{2} ω^{2}}{r^{2} ω^{2}}$
$v^{2} = \frac{- a^{2}}{ω^{2}} + r^{2} ω^{2}$
On comparing equation (i) with the general equation of straight line $y = mx + c$ , we note that the equation (i) is a straight line between $v^{2}$ and $a^{2}$ with negative slope $(\frac{1}{ω^{2}})$
So, option (d) is correct.

CG PET-22.05.2022

Oscillations

140002 A particle of mass $m$ is executing oscillations about the origin on the $x$ -axis. Its potential energy is
$U (x) = k | x |^{3}$ , where $k$ is a positive constant. If the amplitude of oscillation is a, then the time period $T$ is .......

1 Proportional to

\frac{1}{\sqrt{a}}

2 Independent of a

3 Proportional to

\sqrt{a}

4 Proportional to

a^{3 / 2}

Explanation:

A Given that,
$U (x) = k | x |^{3},$
We know that,
$[k] = \frac{[U]}{[x]^{3}} = \frac{[{ML}^{2} T^{- 2}]}{[L^{3}]} = [{ML}^{- 1} T^{- 2}]$
Hence,
The time period is depends on
$T \propto (mass)^{x} (amplitude)^{y} (k)^{z}$
$[{ML}^{0} T^{1}] = [M]^{x} [L]^{y} {[{ML}^{- 1} T^{- 2}]}^{z}$
$[M^{0} L^{0} T^{1}] = [M]^{x + z} [L]^{y - z} [T]^{- 2 z}$
On equating both the side we get -
$- 2 z = 1$
$z = - \frac{1}{2}$
$y - z = 0$
$y = z$
$∴ y = - 1 / 2$
We know that,
$T \propto (amplitude)^{y}$
$T \propto (a)^{y}$
Putting the value of we get -
$T \propto (a)^{- 1 / 2}$
$T \propto \frac{1}{\sqrt{a}}$

CG PET-22.05.2022

Oscillations

139997 For particle $P$ revolving round the centre $O$ with radius of circular path $r$ and angular velocity $ω$ as shown in below figure, the projection of OP on the $x$ -axis at time $t$ is

1

x (t) = r \cos (ω t + \frac{π}{6})

2

x (t) = r \cos (ω t)

3

x (t) = r \sin (ω t + \frac{π}{6})

4

x (t) = r \cos (ω t - \frac{π}{6} ω)

Explanation:

A We know that in polar coordinate the horizontal component
$x = r \cos θ$
But given here $θ = (ω t + ϕ) = 30^{\circ}$
Therefore, putting the value of $θ$ in above equation

$x (t) = r \cos (ω t + 30^{\circ})$
$x (t) = r \cos (ω t + π / 6)$
Hence, projection of OP on $x$ -axis $= r \cos (ω t + π / 6)$

JEE Main-08.04.2023

Oscillations

139999 In a linear simple harmonic motion (SHM)

1 (A), (B) and (C) only

2 (C) and (D) only

3 (A), (B) and (D) only

4 (A), (C) and (D) only

Explanation:

A
We know that, the restoring force in simple harmonic motion
$∵ F = - kx$
- And acceleration in SHM $a = ω^{2} x$
- Velocity is maximum at mean position
- Acceleration is maximum at extreme points

JEE Main-15.04.2023

Oscillations

140000 The displacement of a particle executing SHM is given by $x = 3 \sin [2 π t + \frac{π}{4}]$ where ' $x$ ' is in meters and ' $t$ ' is in seconds. The amplitude and maximum speed of the particle is

1

3 m, 6 π {ms}^{- 1}

2

3 m, 8 π {ms}^{- 1}

3

3 m, 2 π {ms}^{- 1}

4

3 m, 4 π {ms}^{- 1}

Explanation:

A Given,
$x = 3 \sin [2 π t + \frac{π}{4}]$
As we know that the standard equation of SHM is -
$x = A \sin [ω t + ϕ]$
On comparing equation (i) and (ii), we get -
$A = 3 m, ω = 2 π, ϕ = \frac{π}{4}$
$∵ v_{max} = A ω$
$v_{max} = 3 \times 2 π$
$v_{max} = 6 π {ms}^{- 1}$
The amplitude and maximum speed of particle is $3 m$ , $6 π {ms}^{- 1}$ respectively.

Karnataka CET-2022

Oscillations

140001 A mass $m$ is performing linear simple harmonic motion, then which of the following graph represents correctly the variation of acceleration 'a' corresponding to linear velocity 'v' ?

1

2

3

4

Explanation:

D In simple harmonic motion,
$y = r \sin ω t$
$and v = r ω \cos ω t \Rightarrow \cos ω t = \frac{v}{r ω}$
$a = \frac{d v}{a t} = \frac{d}{d t} (r ω \cos ω t)$
$a = - ω^{2} r \sin ω t$
$\sin ω t = - \frac{a}{ω^{2} r}$
Hence,
$\sin^{2} ω t + \cos^{2} ω t = {(- \frac{a}{ω^{2} r})}^{2} + {(\frac{v}{r ω})}^{2}$
$1 = \frac{a^{2}}{ω^{4} r^{2}} + \frac{v^{2}}{r^{2} ω^{2}} [∵ \sin^{2} θ + \cos^{2} θ = 1]$
$- \frac{a^{2}}{ω^{4} r^{2}} = \frac{v^{2}}{r^{2} ω^{2}} - 1$
$- \frac{a^{2}}{ω^{4} r^{2}} = \frac{v^{2} - r^{2} ω^{2}}{r^{2} ω^{2}}$
$v^{2} = \frac{- a^{2}}{ω^{2}} + r^{2} ω^{2}$
On comparing equation (i) with the general equation of straight line $y = mx + c$ , we note that the equation (i) is a straight line between $v^{2}$ and $a^{2}$ with negative slope $(\frac{1}{ω^{2}})$
So, option (d) is correct.

CG PET-22.05.2022

Oscillations

140002 A particle of mass $m$ is executing oscillations about the origin on the $x$ -axis. Its potential energy is
$U (x) = k | x |^{3}$ , where $k$ is a positive constant. If the amplitude of oscillation is a, then the time period $T$ is .......

1 Proportional to

\frac{1}{\sqrt{a}}

2 Independent of a

3 Proportional to

\sqrt{a}

4 Proportional to

a^{3 / 2}

Explanation:

A Given that,
$U (x) = k | x |^{3},$
We know that,
$[k] = \frac{[U]}{[x]^{3}} = \frac{[{ML}^{2} T^{- 2}]}{[L^{3}]} = [{ML}^{- 1} T^{- 2}]$
Hence,
The time period is depends on
$T \propto (mass)^{x} (amplitude)^{y} (k)^{z}$
$[{ML}^{0} T^{1}] = [M]^{x} [L]^{y} {[{ML}^{- 1} T^{- 2}]}^{z}$
$[M^{0} L^{0} T^{1}] = [M]^{x + z} [L]^{y - z} [T]^{- 2 z}$
On equating both the side we get -
$- 2 z = 1$
$z = - \frac{1}{2}$
$y - z = 0$
$y = z$
$∴ y = - 1 / 2$
We know that,
$T \propto (amplitude)^{y}$
$T \propto (a)^{y}$
Putting the value of we get -
$T \propto (a)^{- 1 / 2}$
$T \propto \frac{1}{\sqrt{a}}$

CG PET-22.05.2022

Oscillations

139997 For particle $P$ revolving round the centre $O$ with radius of circular path $r$ and angular velocity $ω$ as shown in below figure, the projection of OP on the $x$ -axis at time $t$ is

1

x (t) = r \cos (ω t + \frac{π}{6})

2

x (t) = r \cos (ω t)

3

x (t) = r \sin (ω t + \frac{π}{6})

4

x (t) = r \cos (ω t - \frac{π}{6} ω)

Explanation:

A We know that in polar coordinate the horizontal component
$x = r \cos θ$
But given here $θ = (ω t + ϕ) = 30^{\circ}$
Therefore, putting the value of $θ$ in above equation

$x (t) = r \cos (ω t + 30^{\circ})$
$x (t) = r \cos (ω t + π / 6)$
Hence, projection of OP on $x$ -axis $= r \cos (ω t + π / 6)$

JEE Main-08.04.2023

Oscillations

139999 In a linear simple harmonic motion (SHM)

1 (A), (B) and (C) only

2 (C) and (D) only

3 (A), (B) and (D) only

4 (A), (C) and (D) only

Explanation:

A
We know that, the restoring force in simple harmonic motion
$∵ F = - kx$
- And acceleration in SHM $a = ω^{2} x$
- Velocity is maximum at mean position
- Acceleration is maximum at extreme points

JEE Main-15.04.2023

Oscillations

140000 The displacement of a particle executing SHM is given by $x = 3 \sin [2 π t + \frac{π}{4}]$ where ' $x$ ' is in meters and ' $t$ ' is in seconds. The amplitude and maximum speed of the particle is

1

3 m, 6 π {ms}^{- 1}

2

3 m, 8 π {ms}^{- 1}

3

3 m, 2 π {ms}^{- 1}

4

3 m, 4 π {ms}^{- 1}

Explanation:

A Given,
$x = 3 \sin [2 π t + \frac{π}{4}]$
As we know that the standard equation of SHM is -
$x = A \sin [ω t + ϕ]$
On comparing equation (i) and (ii), we get -
$A = 3 m, ω = 2 π, ϕ = \frac{π}{4}$
$∵ v_{max} = A ω$
$v_{max} = 3 \times 2 π$
$v_{max} = 6 π {ms}^{- 1}$
The amplitude and maximum speed of particle is $3 m$ , $6 π {ms}^{- 1}$ respectively.

Karnataka CET-2022

Oscillations

140001 A mass $m$ is performing linear simple harmonic motion, then which of the following graph represents correctly the variation of acceleration 'a' corresponding to linear velocity 'v' ?

1

2

3

4

Explanation:

D In simple harmonic motion,
$y = r \sin ω t$
$and v = r ω \cos ω t \Rightarrow \cos ω t = \frac{v}{r ω}$
$a = \frac{d v}{a t} = \frac{d}{d t} (r ω \cos ω t)$
$a = - ω^{2} r \sin ω t$
$\sin ω t = - \frac{a}{ω^{2} r}$
Hence,
$\sin^{2} ω t + \cos^{2} ω t = {(- \frac{a}{ω^{2} r})}^{2} + {(\frac{v}{r ω})}^{2}$
$1 = \frac{a^{2}}{ω^{4} r^{2}} + \frac{v^{2}}{r^{2} ω^{2}} [∵ \sin^{2} θ + \cos^{2} θ = 1]$
$- \frac{a^{2}}{ω^{4} r^{2}} = \frac{v^{2}}{r^{2} ω^{2}} - 1$
$- \frac{a^{2}}{ω^{4} r^{2}} = \frac{v^{2} - r^{2} ω^{2}}{r^{2} ω^{2}}$
$v^{2} = \frac{- a^{2}}{ω^{2}} + r^{2} ω^{2}$
On comparing equation (i) with the general equation of straight line $y = mx + c$ , we note that the equation (i) is a straight line between $v^{2}$ and $a^{2}$ with negative slope $(\frac{1}{ω^{2}})$
So, option (d) is correct.

CG PET-22.05.2022

Oscillations

140002 A particle of mass $m$ is executing oscillations about the origin on the $x$ -axis. Its potential energy is
$U (x) = k | x |^{3}$ , where $k$ is a positive constant. If the amplitude of oscillation is a, then the time period $T$ is .......

1 Proportional to

\frac{1}{\sqrt{a}}

2 Independent of a

3 Proportional to

\sqrt{a}

4 Proportional to

a^{3 / 2}

Explanation:

A Given that,
$U (x) = k | x |^{3},$
We know that,
$[k] = \frac{[U]}{[x]^{3}} = \frac{[{ML}^{2} T^{- 2}]}{[L^{3}]} = [{ML}^{- 1} T^{- 2}]$
Hence,
The time period is depends on
$T \propto (mass)^{x} (amplitude)^{y} (k)^{z}$
$[{ML}^{0} T^{1}] = [M]^{x} [L]^{y} {[{ML}^{- 1} T^{- 2}]}^{z}$
$[M^{0} L^{0} T^{1}] = [M]^{x + z} [L]^{y - z} [T]^{- 2 z}$
On equating both the side we get -
$- 2 z = 1$
$z = - \frac{1}{2}$
$y - z = 0$
$y = z$
$∴ y = - 1 / 2$
We know that,
$T \propto (amplitude)^{y}$
$T \propto (a)^{y}$
Putting the value of we get -
$T \propto (a)^{- 1 / 2}$
$T \propto \frac{1}{\sqrt{a}}$

CG PET-22.05.2022

Oscillations

139997 For particle $P$ revolving round the centre $O$ with radius of circular path $r$ and angular velocity $ω$ as shown in below figure, the projection of OP on the $x$ -axis at time $t$ is

1

x (t) = r \cos (ω t + \frac{π}{6})

2

x (t) = r \cos (ω t)

3

x (t) = r \sin (ω t + \frac{π}{6})

4

x (t) = r \cos (ω t - \frac{π}{6} ω)

Explanation:

A We know that in polar coordinate the horizontal component
$x = r \cos θ$
But given here $θ = (ω t + ϕ) = 30^{\circ}$
Therefore, putting the value of $θ$ in above equation

$x (t) = r \cos (ω t + 30^{\circ})$
$x (t) = r \cos (ω t + π / 6)$
Hence, projection of OP on $x$ -axis $= r \cos (ω t + π / 6)$

JEE Main-08.04.2023

Oscillations

139999 In a linear simple harmonic motion (SHM)

1 (A), (B) and (C) only

2 (C) and (D) only

3 (A), (B) and (D) only

4 (A), (C) and (D) only

Explanation:

A
We know that, the restoring force in simple harmonic motion
$∵ F = - kx$
- And acceleration in SHM $a = ω^{2} x$
- Velocity is maximum at mean position
- Acceleration is maximum at extreme points

JEE Main-15.04.2023

Oscillations

140000 The displacement of a particle executing SHM is given by $x = 3 \sin [2 π t + \frac{π}{4}]$ where ' $x$ ' is in meters and ' $t$ ' is in seconds. The amplitude and maximum speed of the particle is

1

3 m, 6 π {ms}^{- 1}

2

3 m, 8 π {ms}^{- 1}

3

3 m, 2 π {ms}^{- 1}

4

3 m, 4 π {ms}^{- 1}

Explanation:

A Given,
$x = 3 \sin [2 π t + \frac{π}{4}]$
As we know that the standard equation of SHM is -
$x = A \sin [ω t + ϕ]$
On comparing equation (i) and (ii), we get -
$A = 3 m, ω = 2 π, ϕ = \frac{π}{4}$
$∵ v_{max} = A ω$
$v_{max} = 3 \times 2 π$
$v_{max} = 6 π {ms}^{- 1}$
The amplitude and maximum speed of particle is $3 m$ , $6 π {ms}^{- 1}$ respectively.

Karnataka CET-2022

Oscillations

140001 A mass $m$ is performing linear simple harmonic motion, then which of the following graph represents correctly the variation of acceleration 'a' corresponding to linear velocity 'v' ?

1

2

3

4

Explanation:

D In simple harmonic motion,
$y = r \sin ω t$
$and v = r ω \cos ω t \Rightarrow \cos ω t = \frac{v}{r ω}$
$a = \frac{d v}{a t} = \frac{d}{d t} (r ω \cos ω t)$
$a = - ω^{2} r \sin ω t$
$\sin ω t = - \frac{a}{ω^{2} r}$
Hence,
$\sin^{2} ω t + \cos^{2} ω t = {(- \frac{a}{ω^{2} r})}^{2} + {(\frac{v}{r ω})}^{2}$
$1 = \frac{a^{2}}{ω^{4} r^{2}} + \frac{v^{2}}{r^{2} ω^{2}} [∵ \sin^{2} θ + \cos^{2} θ = 1]$
$- \frac{a^{2}}{ω^{4} r^{2}} = \frac{v^{2}}{r^{2} ω^{2}} - 1$
$- \frac{a^{2}}{ω^{4} r^{2}} = \frac{v^{2} - r^{2} ω^{2}}{r^{2} ω^{2}}$
$v^{2} = \frac{- a^{2}}{ω^{2}} + r^{2} ω^{2}$
On comparing equation (i) with the general equation of straight line $y = mx + c$ , we note that the equation (i) is a straight line between $v^{2}$ and $a^{2}$ with negative slope $(\frac{1}{ω^{2}})$
So, option (d) is correct.

CG PET-22.05.2022

Oscillations

140002 A particle of mass $m$ is executing oscillations about the origin on the $x$ -axis. Its potential energy is
$U (x) = k | x |^{3}$ , where $k$ is a positive constant. If the amplitude of oscillation is a, then the time period $T$ is .......

1 Proportional to

\frac{1}{\sqrt{a}}

2 Independent of a

3 Proportional to

\sqrt{a}

4 Proportional to

a^{3 / 2}

Explanation:

A Given that,
$U (x) = k | x |^{3},$
We know that,
$[k] = \frac{[U]}{[x]^{3}} = \frac{[{ML}^{2} T^{- 2}]}{[L^{3}]} = [{ML}^{- 1} T^{- 2}]$
Hence,
The time period is depends on
$T \propto (mass)^{x} (amplitude)^{y} (k)^{z}$
$[{ML}^{0} T^{1}] = [M]^{x} [L]^{y} {[{ML}^{- 1} T^{- 2}]}^{z}$
$[M^{0} L^{0} T^{1}] = [M]^{x + z} [L]^{y - z} [T]^{- 2 z}$
On equating both the side we get -
$- 2 z = 1$
$z = - \frac{1}{2}$
$y - z = 0$
$y = z$
$∴ y = - 1 / 2$
We know that,
$T \propto (amplitude)^{y}$
$T \propto (a)^{y}$
Putting the value of we get -
$T \propto (a)^{- 1 / 2}$
$T \propto \frac{1}{\sqrt{a}}$

CG PET-22.05.2022

Oscillations

139997 For particle $P$ revolving round the centre $O$ with radius of circular path $r$ and angular velocity $ω$ as shown in below figure, the projection of OP on the $x$ -axis at time $t$ is

1

x (t) = r \cos (ω t + \frac{π}{6})

2

x (t) = r \cos (ω t)

3

x (t) = r \sin (ω t + \frac{π}{6})

4

x (t) = r \cos (ω t - \frac{π}{6} ω)

Explanation:

A We know that in polar coordinate the horizontal component
$x = r \cos θ$
But given here $θ = (ω t + ϕ) = 30^{\circ}$
Therefore, putting the value of $θ$ in above equation

$x (t) = r \cos (ω t + 30^{\circ})$
$x (t) = r \cos (ω t + π / 6)$
Hence, projection of OP on $x$ -axis $= r \cos (ω t + π / 6)$

JEE Main-08.04.2023

Oscillations

139999 In a linear simple harmonic motion (SHM)

1 (A), (B) and (C) only

2 (C) and (D) only

3 (A), (B) and (D) only

4 (A), (C) and (D) only

Explanation:

A
We know that, the restoring force in simple harmonic motion
$∵ F = - kx$
- And acceleration in SHM $a = ω^{2} x$
- Velocity is maximum at mean position
- Acceleration is maximum at extreme points

JEE Main-15.04.2023

Oscillations

140000 The displacement of a particle executing SHM is given by $x = 3 \sin [2 π t + \frac{π}{4}]$ where ' $x$ ' is in meters and ' $t$ ' is in seconds. The amplitude and maximum speed of the particle is

1

3 m, 6 π {ms}^{- 1}

2

3 m, 8 π {ms}^{- 1}

3

3 m, 2 π {ms}^{- 1}

4

3 m, 4 π {ms}^{- 1}

Explanation:

A Given,
$x = 3 \sin [2 π t + \frac{π}{4}]$
As we know that the standard equation of SHM is -
$x = A \sin [ω t + ϕ]$
On comparing equation (i) and (ii), we get -
$A = 3 m, ω = 2 π, ϕ = \frac{π}{4}$
$∵ v_{max} = A ω$
$v_{max} = 3 \times 2 π$
$v_{max} = 6 π {ms}^{- 1}$
The amplitude and maximum speed of particle is $3 m$ , $6 π {ms}^{- 1}$ respectively.

Karnataka CET-2022

Oscillations

140001 A mass $m$ is performing linear simple harmonic motion, then which of the following graph represents correctly the variation of acceleration 'a' corresponding to linear velocity 'v' ?

1

2

3

4

Explanation:

D In simple harmonic motion,
$y = r \sin ω t$
$and v = r ω \cos ω t \Rightarrow \cos ω t = \frac{v}{r ω}$
$a = \frac{d v}{a t} = \frac{d}{d t} (r ω \cos ω t)$
$a = - ω^{2} r \sin ω t$
$\sin ω t = - \frac{a}{ω^{2} r}$
Hence,
$\sin^{2} ω t + \cos^{2} ω t = {(- \frac{a}{ω^{2} r})}^{2} + {(\frac{v}{r ω})}^{2}$
$1 = \frac{a^{2}}{ω^{4} r^{2}} + \frac{v^{2}}{r^{2} ω^{2}} [∵ \sin^{2} θ + \cos^{2} θ = 1]$
$- \frac{a^{2}}{ω^{4} r^{2}} = \frac{v^{2}}{r^{2} ω^{2}} - 1$
$- \frac{a^{2}}{ω^{4} r^{2}} = \frac{v^{2} - r^{2} ω^{2}}{r^{2} ω^{2}}$
$v^{2} = \frac{- a^{2}}{ω^{2}} + r^{2} ω^{2}$
On comparing equation (i) with the general equation of straight line $y = mx + c$ , we note that the equation (i) is a straight line between $v^{2}$ and $a^{2}$ with negative slope $(\frac{1}{ω^{2}})$
So, option (d) is correct.

CG PET-22.05.2022

Oscillations

140002 A particle of mass $m$ is executing oscillations about the origin on the $x$ -axis. Its potential energy is
$U (x) = k | x |^{3}$ , where $k$ is a positive constant. If the amplitude of oscillation is a, then the time period $T$ is .......

1 Proportional to

\frac{1}{\sqrt{a}}

2 Independent of a

3 Proportional to

\sqrt{a}

4 Proportional to

a^{3 / 2}

Explanation:

A Given that,
$U (x) = k | x |^{3},$
We know that,
$[k] = \frac{[U]}{[x]^{3}} = \frac{[{ML}^{2} T^{- 2}]}{[L^{3}]} = [{ML}^{- 1} T^{- 2}]$
Hence,
The time period is depends on
$T \propto (mass)^{x} (amplitude)^{y} (k)^{z}$
$[{ML}^{0} T^{1}] = [M]^{x} [L]^{y} {[{ML}^{- 1} T^{- 2}]}^{z}$
$[M^{0} L^{0} T^{1}] = [M]^{x + z} [L]^{y - z} [T]^{- 2 z}$
On equating both the side we get -
$- 2 z = 1$
$z = - \frac{1}{2}$
$y - z = 0$
$y = z$
$∴ y = - 1 / 2$
We know that,
$T \propto (amplitude)^{y}$
$T \propto (a)^{y}$
Putting the value of we get -
$T \propto (a)^{- 1 / 2}$
$T \propto \frac{1}{\sqrt{a}}$

CG PET-22.05.2022