139997
For particle $P$ revolving round the centre $O$ with radius of circular path $\mathbf{r}$ and angular velocity $\omega$ as shown in below figure, the projection of OP on the $x$-axis at time $t$ is
A We know that in polar coordinate the horizontal component $\mathrm{x}=\mathrm{r} \cos \theta$ But given here $\theta=(\omega \mathrm{t}+\phi)=30^{\circ}$ Therefore, putting the value of $\theta$ in above equation $x(t)=r \cos \left(\omega t+30^{\circ}\right)$ $x(t)=r \cos (\omega t+\pi / 6)$ Hence, projection of OP on $\mathrm{x}$-axis $=\mathrm{r} \cos (\omega \mathrm{t}+\pi / 6)$
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 $\because \quad \mathrm{F}=-\mathrm{kx}$ - And acceleration in SHM $a=\omega^{2} x$ - 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 $x=3 \sin \left[2 \pi t+\frac{\pi}{4}\right]$ where ' $x$ ' is in meters and ' $t$ ' is in seconds. The amplitude and maximum speed of the particle is
1 $3 \mathrm{~m}, 6 \pi \mathrm{ms}^{-1}$
2 $3 \mathrm{~m}, 8 \pi \mathrm{ms}^{-1}$
3 $3 \mathrm{~m}, 2 \pi \mathrm{ms}^{-1}$
4 $3 \mathrm{~m}, 4 \pi \mathrm{ms}^{-1}$
Explanation:
A Given, $x=3 \sin \left[2 \pi t+\frac{\pi}{4}\right]$ As we know that the standard equation of SHM is - $\mathrm{x}=\mathrm{A} \sin [\omega \mathrm{t}+\phi]$ On comparing equation (i) and (ii), we get - $\mathrm{A}=3 \mathrm{~m}, \omega=2 \pi, \phi=\frac{\pi}{4}$ $\because \quad \mathrm{v}_{\max }=\mathrm{A} \omega$ $\mathrm{v}_{\max }=3 \times 2 \pi$ $\mathrm{v}_{\max }=6 \pi \mathrm{ms}^{-1}$ The amplitude and maximum speed of particle is $3 \mathrm{~m}$, $6 \pi \mathrm{ms}^{-1}$ 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, $\mathrm{y}=\mathrm{r} \sin \omega \mathrm{t}$ $\text { and } \quad \mathrm{v}=\mathrm{r} \omega \cos \omega \mathrm{t} \Rightarrow \cos \omega \mathrm{t}=\frac{\mathrm{v}}{\mathrm{r} \omega}$ $a=\frac{d v}{a t}=\frac{d}{d t}(r \omega \cos \omega t)$ $a=-\omega^{2} r \sin \omega t$ $\sin \omega t=-\frac{a}{\omega^{2} r}$ Hence, $\sin ^{2} \omega t+\cos ^{2} \omega t=\left(-\frac{a}{\omega^{2} r}\right)^{2}+\left(\frac{v}{r \omega}\right)^{2}$ $1=\frac{a^{2}}{\omega^{4} r^{2}}+\frac{v^{2}}{r^{2} \omega^{2}} \quad\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$ $-\frac{a^{2}}{\omega^{4} r^{2}}=\frac{v^{2}}{r^{2} \omega^{2}}-1$ $-\frac{a^{2}}{\omega^{4} r^{2}}=\frac{v^{2}-r^{2} \omega^{2}}{r^{2} \omega^{2}}$ $v^{2}=\frac{-a^{2}}{\omega^{2}}+r^{2} \omega^{2}$ On comparing equation (i) with the general equation of straight line $\mathrm{y}=\mathrm{mx}+\mathrm{c}$, we note that the equation (i) is a straight line between $v^{2}$ and $a^{2}$ with negative slope $\left(\frac{1}{\omega^{2}}\right)$ 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 $\mathrm{U}(\mathrm{x})=\mathrm{k}|\mathrm{x}|^{3}$, where $\mathrm{k}$ is a positive constant. If the amplitude of oscillation is a, then the time period $T$ is .......
1 Proportional to $\frac{1}{\sqrt{\mathrm{a}}}$
2 Independent of a
3 Proportional to $\sqrt{\mathrm{a}}$
4 Proportional to $\mathrm{a}^{3 / 2}$
Explanation:
A Given that, $\mathrm{U}(\mathrm{x})=\mathrm{k}|\mathrm{x}|^{3} \text {, }$ We know that, $[\mathrm{k}]=\frac{[\mathrm{U}]}{[\mathrm{x}]^{3}}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^{3}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$ Hence, The time period is depends on $\mathrm{T} \propto(\text { mass })^{\mathrm{x}}(\text { amplitude })^{\mathrm{y}}(\mathrm{k})^{\mathrm{z}}$ ${\left[\mathrm{ML}^{0} \mathrm{~T}^{1}\right]=[\mathrm{M}]^{\mathrm{x}}[\mathrm{L}]^{\mathrm{y}}\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]^{\mathrm{z}}}$ ${\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1}\right]=[\mathrm{M}]^{\mathrm{x}+\mathrm{z}}[\mathrm{L}]^{\mathrm{y}-\mathrm{z}}[\mathrm{T}]^{-2 \mathrm{z}}}$ On equating both the side we get - $-2 \mathrm{z}=1$ $\mathrm{z}=-\frac{1}{2}$ $\mathrm{y}-\mathrm{z}=0$ $\mathrm{y} =\mathrm{z}$ $\therefore \quad \mathrm{y} =-1 / 2$ We know that, $\mathrm{T} \propto(\text { amplitude })^{\mathrm{y}}$ $\mathrm{T} \propto(\mathrm{a})^{\mathrm{y}}$ Putting the value of we get - $\mathrm{T} \propto(\mathrm{a})^{-1 / 2}$ $\mathrm{~T} \propto \frac{1}{\sqrt{\mathrm{a}}}$
139997
For particle $P$ revolving round the centre $O$ with radius of circular path $\mathbf{r}$ and angular velocity $\omega$ as shown in below figure, the projection of OP on the $x$-axis at time $t$ is
A We know that in polar coordinate the horizontal component $\mathrm{x}=\mathrm{r} \cos \theta$ But given here $\theta=(\omega \mathrm{t}+\phi)=30^{\circ}$ Therefore, putting the value of $\theta$ in above equation $x(t)=r \cos \left(\omega t+30^{\circ}\right)$ $x(t)=r \cos (\omega t+\pi / 6)$ Hence, projection of OP on $\mathrm{x}$-axis $=\mathrm{r} \cos (\omega \mathrm{t}+\pi / 6)$
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 $\because \quad \mathrm{F}=-\mathrm{kx}$ - And acceleration in SHM $a=\omega^{2} x$ - 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 $x=3 \sin \left[2 \pi t+\frac{\pi}{4}\right]$ where ' $x$ ' is in meters and ' $t$ ' is in seconds. The amplitude and maximum speed of the particle is
1 $3 \mathrm{~m}, 6 \pi \mathrm{ms}^{-1}$
2 $3 \mathrm{~m}, 8 \pi \mathrm{ms}^{-1}$
3 $3 \mathrm{~m}, 2 \pi \mathrm{ms}^{-1}$
4 $3 \mathrm{~m}, 4 \pi \mathrm{ms}^{-1}$
Explanation:
A Given, $x=3 \sin \left[2 \pi t+\frac{\pi}{4}\right]$ As we know that the standard equation of SHM is - $\mathrm{x}=\mathrm{A} \sin [\omega \mathrm{t}+\phi]$ On comparing equation (i) and (ii), we get - $\mathrm{A}=3 \mathrm{~m}, \omega=2 \pi, \phi=\frac{\pi}{4}$ $\because \quad \mathrm{v}_{\max }=\mathrm{A} \omega$ $\mathrm{v}_{\max }=3 \times 2 \pi$ $\mathrm{v}_{\max }=6 \pi \mathrm{ms}^{-1}$ The amplitude and maximum speed of particle is $3 \mathrm{~m}$, $6 \pi \mathrm{ms}^{-1}$ 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, $\mathrm{y}=\mathrm{r} \sin \omega \mathrm{t}$ $\text { and } \quad \mathrm{v}=\mathrm{r} \omega \cos \omega \mathrm{t} \Rightarrow \cos \omega \mathrm{t}=\frac{\mathrm{v}}{\mathrm{r} \omega}$ $a=\frac{d v}{a t}=\frac{d}{d t}(r \omega \cos \omega t)$ $a=-\omega^{2} r \sin \omega t$ $\sin \omega t=-\frac{a}{\omega^{2} r}$ Hence, $\sin ^{2} \omega t+\cos ^{2} \omega t=\left(-\frac{a}{\omega^{2} r}\right)^{2}+\left(\frac{v}{r \omega}\right)^{2}$ $1=\frac{a^{2}}{\omega^{4} r^{2}}+\frac{v^{2}}{r^{2} \omega^{2}} \quad\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$ $-\frac{a^{2}}{\omega^{4} r^{2}}=\frac{v^{2}}{r^{2} \omega^{2}}-1$ $-\frac{a^{2}}{\omega^{4} r^{2}}=\frac{v^{2}-r^{2} \omega^{2}}{r^{2} \omega^{2}}$ $v^{2}=\frac{-a^{2}}{\omega^{2}}+r^{2} \omega^{2}$ On comparing equation (i) with the general equation of straight line $\mathrm{y}=\mathrm{mx}+\mathrm{c}$, we note that the equation (i) is a straight line between $v^{2}$ and $a^{2}$ with negative slope $\left(\frac{1}{\omega^{2}}\right)$ 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 $\mathrm{U}(\mathrm{x})=\mathrm{k}|\mathrm{x}|^{3}$, where $\mathrm{k}$ is a positive constant. If the amplitude of oscillation is a, then the time period $T$ is .......
1 Proportional to $\frac{1}{\sqrt{\mathrm{a}}}$
2 Independent of a
3 Proportional to $\sqrt{\mathrm{a}}$
4 Proportional to $\mathrm{a}^{3 / 2}$
Explanation:
A Given that, $\mathrm{U}(\mathrm{x})=\mathrm{k}|\mathrm{x}|^{3} \text {, }$ We know that, $[\mathrm{k}]=\frac{[\mathrm{U}]}{[\mathrm{x}]^{3}}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^{3}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$ Hence, The time period is depends on $\mathrm{T} \propto(\text { mass })^{\mathrm{x}}(\text { amplitude })^{\mathrm{y}}(\mathrm{k})^{\mathrm{z}}$ ${\left[\mathrm{ML}^{0} \mathrm{~T}^{1}\right]=[\mathrm{M}]^{\mathrm{x}}[\mathrm{L}]^{\mathrm{y}}\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]^{\mathrm{z}}}$ ${\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1}\right]=[\mathrm{M}]^{\mathrm{x}+\mathrm{z}}[\mathrm{L}]^{\mathrm{y}-\mathrm{z}}[\mathrm{T}]^{-2 \mathrm{z}}}$ On equating both the side we get - $-2 \mathrm{z}=1$ $\mathrm{z}=-\frac{1}{2}$ $\mathrm{y}-\mathrm{z}=0$ $\mathrm{y} =\mathrm{z}$ $\therefore \quad \mathrm{y} =-1 / 2$ We know that, $\mathrm{T} \propto(\text { amplitude })^{\mathrm{y}}$ $\mathrm{T} \propto(\mathrm{a})^{\mathrm{y}}$ Putting the value of we get - $\mathrm{T} \propto(\mathrm{a})^{-1 / 2}$ $\mathrm{~T} \propto \frac{1}{\sqrt{\mathrm{a}}}$
139997
For particle $P$ revolving round the centre $O$ with radius of circular path $\mathbf{r}$ and angular velocity $\omega$ as shown in below figure, the projection of OP on the $x$-axis at time $t$ is
A We know that in polar coordinate the horizontal component $\mathrm{x}=\mathrm{r} \cos \theta$ But given here $\theta=(\omega \mathrm{t}+\phi)=30^{\circ}$ Therefore, putting the value of $\theta$ in above equation $x(t)=r \cos \left(\omega t+30^{\circ}\right)$ $x(t)=r \cos (\omega t+\pi / 6)$ Hence, projection of OP on $\mathrm{x}$-axis $=\mathrm{r} \cos (\omega \mathrm{t}+\pi / 6)$
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 $\because \quad \mathrm{F}=-\mathrm{kx}$ - And acceleration in SHM $a=\omega^{2} x$ - 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 $x=3 \sin \left[2 \pi t+\frac{\pi}{4}\right]$ where ' $x$ ' is in meters and ' $t$ ' is in seconds. The amplitude and maximum speed of the particle is
1 $3 \mathrm{~m}, 6 \pi \mathrm{ms}^{-1}$
2 $3 \mathrm{~m}, 8 \pi \mathrm{ms}^{-1}$
3 $3 \mathrm{~m}, 2 \pi \mathrm{ms}^{-1}$
4 $3 \mathrm{~m}, 4 \pi \mathrm{ms}^{-1}$
Explanation:
A Given, $x=3 \sin \left[2 \pi t+\frac{\pi}{4}\right]$ As we know that the standard equation of SHM is - $\mathrm{x}=\mathrm{A} \sin [\omega \mathrm{t}+\phi]$ On comparing equation (i) and (ii), we get - $\mathrm{A}=3 \mathrm{~m}, \omega=2 \pi, \phi=\frac{\pi}{4}$ $\because \quad \mathrm{v}_{\max }=\mathrm{A} \omega$ $\mathrm{v}_{\max }=3 \times 2 \pi$ $\mathrm{v}_{\max }=6 \pi \mathrm{ms}^{-1}$ The amplitude and maximum speed of particle is $3 \mathrm{~m}$, $6 \pi \mathrm{ms}^{-1}$ 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, $\mathrm{y}=\mathrm{r} \sin \omega \mathrm{t}$ $\text { and } \quad \mathrm{v}=\mathrm{r} \omega \cos \omega \mathrm{t} \Rightarrow \cos \omega \mathrm{t}=\frac{\mathrm{v}}{\mathrm{r} \omega}$ $a=\frac{d v}{a t}=\frac{d}{d t}(r \omega \cos \omega t)$ $a=-\omega^{2} r \sin \omega t$ $\sin \omega t=-\frac{a}{\omega^{2} r}$ Hence, $\sin ^{2} \omega t+\cos ^{2} \omega t=\left(-\frac{a}{\omega^{2} r}\right)^{2}+\left(\frac{v}{r \omega}\right)^{2}$ $1=\frac{a^{2}}{\omega^{4} r^{2}}+\frac{v^{2}}{r^{2} \omega^{2}} \quad\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$ $-\frac{a^{2}}{\omega^{4} r^{2}}=\frac{v^{2}}{r^{2} \omega^{2}}-1$ $-\frac{a^{2}}{\omega^{4} r^{2}}=\frac{v^{2}-r^{2} \omega^{2}}{r^{2} \omega^{2}}$ $v^{2}=\frac{-a^{2}}{\omega^{2}}+r^{2} \omega^{2}$ On comparing equation (i) with the general equation of straight line $\mathrm{y}=\mathrm{mx}+\mathrm{c}$, we note that the equation (i) is a straight line between $v^{2}$ and $a^{2}$ with negative slope $\left(\frac{1}{\omega^{2}}\right)$ 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 $\mathrm{U}(\mathrm{x})=\mathrm{k}|\mathrm{x}|^{3}$, where $\mathrm{k}$ is a positive constant. If the amplitude of oscillation is a, then the time period $T$ is .......
1 Proportional to $\frac{1}{\sqrt{\mathrm{a}}}$
2 Independent of a
3 Proportional to $\sqrt{\mathrm{a}}$
4 Proportional to $\mathrm{a}^{3 / 2}$
Explanation:
A Given that, $\mathrm{U}(\mathrm{x})=\mathrm{k}|\mathrm{x}|^{3} \text {, }$ We know that, $[\mathrm{k}]=\frac{[\mathrm{U}]}{[\mathrm{x}]^{3}}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^{3}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$ Hence, The time period is depends on $\mathrm{T} \propto(\text { mass })^{\mathrm{x}}(\text { amplitude })^{\mathrm{y}}(\mathrm{k})^{\mathrm{z}}$ ${\left[\mathrm{ML}^{0} \mathrm{~T}^{1}\right]=[\mathrm{M}]^{\mathrm{x}}[\mathrm{L}]^{\mathrm{y}}\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]^{\mathrm{z}}}$ ${\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1}\right]=[\mathrm{M}]^{\mathrm{x}+\mathrm{z}}[\mathrm{L}]^{\mathrm{y}-\mathrm{z}}[\mathrm{T}]^{-2 \mathrm{z}}}$ On equating both the side we get - $-2 \mathrm{z}=1$ $\mathrm{z}=-\frac{1}{2}$ $\mathrm{y}-\mathrm{z}=0$ $\mathrm{y} =\mathrm{z}$ $\therefore \quad \mathrm{y} =-1 / 2$ We know that, $\mathrm{T} \propto(\text { amplitude })^{\mathrm{y}}$ $\mathrm{T} \propto(\mathrm{a})^{\mathrm{y}}$ Putting the value of we get - $\mathrm{T} \propto(\mathrm{a})^{-1 / 2}$ $\mathrm{~T} \propto \frac{1}{\sqrt{\mathrm{a}}}$
139997
For particle $P$ revolving round the centre $O$ with radius of circular path $\mathbf{r}$ and angular velocity $\omega$ as shown in below figure, the projection of OP on the $x$-axis at time $t$ is
A We know that in polar coordinate the horizontal component $\mathrm{x}=\mathrm{r} \cos \theta$ But given here $\theta=(\omega \mathrm{t}+\phi)=30^{\circ}$ Therefore, putting the value of $\theta$ in above equation $x(t)=r \cos \left(\omega t+30^{\circ}\right)$ $x(t)=r \cos (\omega t+\pi / 6)$ Hence, projection of OP on $\mathrm{x}$-axis $=\mathrm{r} \cos (\omega \mathrm{t}+\pi / 6)$
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 $\because \quad \mathrm{F}=-\mathrm{kx}$ - And acceleration in SHM $a=\omega^{2} x$ - 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 $x=3 \sin \left[2 \pi t+\frac{\pi}{4}\right]$ where ' $x$ ' is in meters and ' $t$ ' is in seconds. The amplitude and maximum speed of the particle is
1 $3 \mathrm{~m}, 6 \pi \mathrm{ms}^{-1}$
2 $3 \mathrm{~m}, 8 \pi \mathrm{ms}^{-1}$
3 $3 \mathrm{~m}, 2 \pi \mathrm{ms}^{-1}$
4 $3 \mathrm{~m}, 4 \pi \mathrm{ms}^{-1}$
Explanation:
A Given, $x=3 \sin \left[2 \pi t+\frac{\pi}{4}\right]$ As we know that the standard equation of SHM is - $\mathrm{x}=\mathrm{A} \sin [\omega \mathrm{t}+\phi]$ On comparing equation (i) and (ii), we get - $\mathrm{A}=3 \mathrm{~m}, \omega=2 \pi, \phi=\frac{\pi}{4}$ $\because \quad \mathrm{v}_{\max }=\mathrm{A} \omega$ $\mathrm{v}_{\max }=3 \times 2 \pi$ $\mathrm{v}_{\max }=6 \pi \mathrm{ms}^{-1}$ The amplitude and maximum speed of particle is $3 \mathrm{~m}$, $6 \pi \mathrm{ms}^{-1}$ 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, $\mathrm{y}=\mathrm{r} \sin \omega \mathrm{t}$ $\text { and } \quad \mathrm{v}=\mathrm{r} \omega \cos \omega \mathrm{t} \Rightarrow \cos \omega \mathrm{t}=\frac{\mathrm{v}}{\mathrm{r} \omega}$ $a=\frac{d v}{a t}=\frac{d}{d t}(r \omega \cos \omega t)$ $a=-\omega^{2} r \sin \omega t$ $\sin \omega t=-\frac{a}{\omega^{2} r}$ Hence, $\sin ^{2} \omega t+\cos ^{2} \omega t=\left(-\frac{a}{\omega^{2} r}\right)^{2}+\left(\frac{v}{r \omega}\right)^{2}$ $1=\frac{a^{2}}{\omega^{4} r^{2}}+\frac{v^{2}}{r^{2} \omega^{2}} \quad\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$ $-\frac{a^{2}}{\omega^{4} r^{2}}=\frac{v^{2}}{r^{2} \omega^{2}}-1$ $-\frac{a^{2}}{\omega^{4} r^{2}}=\frac{v^{2}-r^{2} \omega^{2}}{r^{2} \omega^{2}}$ $v^{2}=\frac{-a^{2}}{\omega^{2}}+r^{2} \omega^{2}$ On comparing equation (i) with the general equation of straight line $\mathrm{y}=\mathrm{mx}+\mathrm{c}$, we note that the equation (i) is a straight line between $v^{2}$ and $a^{2}$ with negative slope $\left(\frac{1}{\omega^{2}}\right)$ 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 $\mathrm{U}(\mathrm{x})=\mathrm{k}|\mathrm{x}|^{3}$, where $\mathrm{k}$ is a positive constant. If the amplitude of oscillation is a, then the time period $T$ is .......
1 Proportional to $\frac{1}{\sqrt{\mathrm{a}}}$
2 Independent of a
3 Proportional to $\sqrt{\mathrm{a}}$
4 Proportional to $\mathrm{a}^{3 / 2}$
Explanation:
A Given that, $\mathrm{U}(\mathrm{x})=\mathrm{k}|\mathrm{x}|^{3} \text {, }$ We know that, $[\mathrm{k}]=\frac{[\mathrm{U}]}{[\mathrm{x}]^{3}}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^{3}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$ Hence, The time period is depends on $\mathrm{T} \propto(\text { mass })^{\mathrm{x}}(\text { amplitude })^{\mathrm{y}}(\mathrm{k})^{\mathrm{z}}$ ${\left[\mathrm{ML}^{0} \mathrm{~T}^{1}\right]=[\mathrm{M}]^{\mathrm{x}}[\mathrm{L}]^{\mathrm{y}}\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]^{\mathrm{z}}}$ ${\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1}\right]=[\mathrm{M}]^{\mathrm{x}+\mathrm{z}}[\mathrm{L}]^{\mathrm{y}-\mathrm{z}}[\mathrm{T}]^{-2 \mathrm{z}}}$ On equating both the side we get - $-2 \mathrm{z}=1$ $\mathrm{z}=-\frac{1}{2}$ $\mathrm{y}-\mathrm{z}=0$ $\mathrm{y} =\mathrm{z}$ $\therefore \quad \mathrm{y} =-1 / 2$ We know that, $\mathrm{T} \propto(\text { amplitude })^{\mathrm{y}}$ $\mathrm{T} \propto(\mathrm{a})^{\mathrm{y}}$ Putting the value of we get - $\mathrm{T} \propto(\mathrm{a})^{-1 / 2}$ $\mathrm{~T} \propto \frac{1}{\sqrt{\mathrm{a}}}$
139997
For particle $P$ revolving round the centre $O$ with radius of circular path $\mathbf{r}$ and angular velocity $\omega$ as shown in below figure, the projection of OP on the $x$-axis at time $t$ is
A We know that in polar coordinate the horizontal component $\mathrm{x}=\mathrm{r} \cos \theta$ But given here $\theta=(\omega \mathrm{t}+\phi)=30^{\circ}$ Therefore, putting the value of $\theta$ in above equation $x(t)=r \cos \left(\omega t+30^{\circ}\right)$ $x(t)=r \cos (\omega t+\pi / 6)$ Hence, projection of OP on $\mathrm{x}$-axis $=\mathrm{r} \cos (\omega \mathrm{t}+\pi / 6)$
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 $\because \quad \mathrm{F}=-\mathrm{kx}$ - And acceleration in SHM $a=\omega^{2} x$ - 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 $x=3 \sin \left[2 \pi t+\frac{\pi}{4}\right]$ where ' $x$ ' is in meters and ' $t$ ' is in seconds. The amplitude and maximum speed of the particle is
1 $3 \mathrm{~m}, 6 \pi \mathrm{ms}^{-1}$
2 $3 \mathrm{~m}, 8 \pi \mathrm{ms}^{-1}$
3 $3 \mathrm{~m}, 2 \pi \mathrm{ms}^{-1}$
4 $3 \mathrm{~m}, 4 \pi \mathrm{ms}^{-1}$
Explanation:
A Given, $x=3 \sin \left[2 \pi t+\frac{\pi}{4}\right]$ As we know that the standard equation of SHM is - $\mathrm{x}=\mathrm{A} \sin [\omega \mathrm{t}+\phi]$ On comparing equation (i) and (ii), we get - $\mathrm{A}=3 \mathrm{~m}, \omega=2 \pi, \phi=\frac{\pi}{4}$ $\because \quad \mathrm{v}_{\max }=\mathrm{A} \omega$ $\mathrm{v}_{\max }=3 \times 2 \pi$ $\mathrm{v}_{\max }=6 \pi \mathrm{ms}^{-1}$ The amplitude and maximum speed of particle is $3 \mathrm{~m}$, $6 \pi \mathrm{ms}^{-1}$ 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, $\mathrm{y}=\mathrm{r} \sin \omega \mathrm{t}$ $\text { and } \quad \mathrm{v}=\mathrm{r} \omega \cos \omega \mathrm{t} \Rightarrow \cos \omega \mathrm{t}=\frac{\mathrm{v}}{\mathrm{r} \omega}$ $a=\frac{d v}{a t}=\frac{d}{d t}(r \omega \cos \omega t)$ $a=-\omega^{2} r \sin \omega t$ $\sin \omega t=-\frac{a}{\omega^{2} r}$ Hence, $\sin ^{2} \omega t+\cos ^{2} \omega t=\left(-\frac{a}{\omega^{2} r}\right)^{2}+\left(\frac{v}{r \omega}\right)^{2}$ $1=\frac{a^{2}}{\omega^{4} r^{2}}+\frac{v^{2}}{r^{2} \omega^{2}} \quad\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$ $-\frac{a^{2}}{\omega^{4} r^{2}}=\frac{v^{2}}{r^{2} \omega^{2}}-1$ $-\frac{a^{2}}{\omega^{4} r^{2}}=\frac{v^{2}-r^{2} \omega^{2}}{r^{2} \omega^{2}}$ $v^{2}=\frac{-a^{2}}{\omega^{2}}+r^{2} \omega^{2}$ On comparing equation (i) with the general equation of straight line $\mathrm{y}=\mathrm{mx}+\mathrm{c}$, we note that the equation (i) is a straight line between $v^{2}$ and $a^{2}$ with negative slope $\left(\frac{1}{\omega^{2}}\right)$ 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 $\mathrm{U}(\mathrm{x})=\mathrm{k}|\mathrm{x}|^{3}$, where $\mathrm{k}$ is a positive constant. If the amplitude of oscillation is a, then the time period $T$ is .......
1 Proportional to $\frac{1}{\sqrt{\mathrm{a}}}$
2 Independent of a
3 Proportional to $\sqrt{\mathrm{a}}$
4 Proportional to $\mathrm{a}^{3 / 2}$
Explanation:
A Given that, $\mathrm{U}(\mathrm{x})=\mathrm{k}|\mathrm{x}|^{3} \text {, }$ We know that, $[\mathrm{k}]=\frac{[\mathrm{U}]}{[\mathrm{x}]^{3}}=\frac{\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]}{\left[\mathrm{L}^{3}\right]}=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$ Hence, The time period is depends on $\mathrm{T} \propto(\text { mass })^{\mathrm{x}}(\text { amplitude })^{\mathrm{y}}(\mathrm{k})^{\mathrm{z}}$ ${\left[\mathrm{ML}^{0} \mathrm{~T}^{1}\right]=[\mathrm{M}]^{\mathrm{x}}[\mathrm{L}]^{\mathrm{y}}\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]^{\mathrm{z}}}$ ${\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{1}\right]=[\mathrm{M}]^{\mathrm{x}+\mathrm{z}}[\mathrm{L}]^{\mathrm{y}-\mathrm{z}}[\mathrm{T}]^{-2 \mathrm{z}}}$ On equating both the side we get - $-2 \mathrm{z}=1$ $\mathrm{z}=-\frac{1}{2}$ $\mathrm{y}-\mathrm{z}=0$ $\mathrm{y} =\mathrm{z}$ $\therefore \quad \mathrm{y} =-1 / 2$ We know that, $\mathrm{T} \propto(\text { amplitude })^{\mathrm{y}}$ $\mathrm{T} \propto(\mathrm{a})^{\mathrm{y}}$ Putting the value of we get - $\mathrm{T} \propto(\mathrm{a})^{-1 / 2}$ $\mathrm{~T} \propto \frac{1}{\sqrt{\mathrm{a}}}$
