140574
Two pendulum oscillate with a constant phase difference of $45^{\circ}$ and same amplitude. If the maximum velocity of one of them is $v$ and that of other is $v+x$, then the value of $x$ will be
1 0
2 $\frac{\mathrm{V}}{2}$
3 $\frac{\mathrm{V}}{\sqrt{2}}$
4 $(\sqrt{2}) \mathrm{v}$
Explanation:
A Let, velocity of one pendulum is $\mathrm{v}_{1}=\mathrm{v}_{\mathrm{o}} \cos \left(\omega \mathrm{t}+\phi_{1}\right)$ Velocity of the other pendulum is- $\mathrm{v}_{2}=\mathrm{v}_{\mathrm{o}} \cos \left(\omega \mathrm{t}+\phi_{2}\right)$ According to the question- $\phi_{2}-\phi_{1}=45^{\circ}$ Clearly, $\left|\mathrm{v}_{1}\right|_{\max }=\left|\mathrm{v}_{2}\right|_{\max }=\mathrm{v}_{\mathrm{o}}=\mathrm{v}$ Thus, $\quad \mathrm{x}=0$
J and K-CET-2015
Oscillations
140576
The period of oscillation of a simple pendulum of length $l$ suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination $\alpha$, is given by
A Given, simple pendulum of length $=l$ Acceleration $=g \sin \theta$. Since vehicle is accelerating a pseudo force $\mathrm{mg} \sin \theta$ will act on bob of pendulum which cancel the $\sin \theta$ component of weight of the bob. Hence, we can say that the effective acceleration would $=\mathrm{g} \cos \alpha$ Now, the time period of oscillation- $\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g} \cos \alpha}}$
JEE Main-29.07.2022
Oscillations
140578
Time period of a simple pendulum of length $l$ is $T_{1}$ and time period of a uniform rod of the same length $l$ pivoted about one end and oscillating in a vertical plane is $T_{2}$. Amplitude of oscillations in both the cases is small. The, $T_{1} / T_{2}$ is
1 $\frac{1}{\sqrt{3}}$
2 1
3 $\sqrt{\frac{4}{3}}$
4 $\sqrt{\frac{3}{2}}$
Explanation:
D The period of simple pendulum is given by $\mathrm{T}_{1}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ Time period of uniform rod at given position - $\mathrm{T}_{2}=2 \pi \sqrt{\frac{\text { Inertia factor }}{\text { Spring factor }}}$ $\because \quad$ Inertia factor $=$ Moment of inertia of rod at one end $=\frac{\mathrm{m} l^{2}}{12}+\frac{\mathrm{m} l^{2}}{4}$ $=\frac{\mathrm{m} l^{2}+3 \mathrm{~m} l^{2}}{12}=\frac{4 \mathrm{~m} l^{2}}{12}=\frac{\mathrm{m} l^{2}}{3}$ And spring factor $=$ Restoring torque per unit angular displacement $=\mathrm{mg} \times \frac{l}{2} \frac{\sin \theta}{\theta}=\mathrm{mg} \frac{l}{2} \quad[\because \theta \text { is very small }]$ $\therefore \quad \mathrm{T}_{2} =2 \pi \sqrt{\frac{\mathrm{m} l^{2} / 3}{\mathrm{mgl} / 2}}$ $\mathrm{~T}_{2} =2 \pi \sqrt{\frac{2 l}{3 \mathrm{~g}}}$ $\text { Hence, } \frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}} =\sqrt{\frac{3}{2}}$
UP CPMT-2006
Oscillations
140579
The angular amplitude of simple pendulum is $\boldsymbol{\theta}_{0}$. The maximum tension in its string will be
D Let, mass of bob $=\mathrm{m}$, string length $=l$ Maximum tension in the string $\left(\mathrm{T}_{\text {max. }}\right)=m g+\frac{m v^{2}}{l}$ $=\mathrm{mg}+\frac{2 \mathrm{mg} l}{l}\left(1-\cos \theta_{0}\right)$ \(=\mathrm{mg}+\frac{2 \mathrm{mg} l}{l}\left(1-\cos \theta_0\right) \quad\left[\begin{array}{rl}\because \cos \theta_0 & =\frac{l-\mathrm{h}}{l} \\ \mathrm{v}^2 & =2 \mathrm{gh}\end{array}\right]\) $=\mathrm{mg}+2 \mathrm{mg}\left(1-1+\frac{\theta_{0}^{2}}{2}\right) \quad\left(\theta_{0} \text { is small }\right)$ $=\operatorname{mg}\left(1+\theta_{0}^{2}\right)$
140574
Two pendulum oscillate with a constant phase difference of $45^{\circ}$ and same amplitude. If the maximum velocity of one of them is $v$ and that of other is $v+x$, then the value of $x$ will be
1 0
2 $\frac{\mathrm{V}}{2}$
3 $\frac{\mathrm{V}}{\sqrt{2}}$
4 $(\sqrt{2}) \mathrm{v}$
Explanation:
A Let, velocity of one pendulum is $\mathrm{v}_{1}=\mathrm{v}_{\mathrm{o}} \cos \left(\omega \mathrm{t}+\phi_{1}\right)$ Velocity of the other pendulum is- $\mathrm{v}_{2}=\mathrm{v}_{\mathrm{o}} \cos \left(\omega \mathrm{t}+\phi_{2}\right)$ According to the question- $\phi_{2}-\phi_{1}=45^{\circ}$ Clearly, $\left|\mathrm{v}_{1}\right|_{\max }=\left|\mathrm{v}_{2}\right|_{\max }=\mathrm{v}_{\mathrm{o}}=\mathrm{v}$ Thus, $\quad \mathrm{x}=0$
J and K-CET-2015
Oscillations
140576
The period of oscillation of a simple pendulum of length $l$ suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination $\alpha$, is given by
A Given, simple pendulum of length $=l$ Acceleration $=g \sin \theta$. Since vehicle is accelerating a pseudo force $\mathrm{mg} \sin \theta$ will act on bob of pendulum which cancel the $\sin \theta$ component of weight of the bob. Hence, we can say that the effective acceleration would $=\mathrm{g} \cos \alpha$ Now, the time period of oscillation- $\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g} \cos \alpha}}$
JEE Main-29.07.2022
Oscillations
140578
Time period of a simple pendulum of length $l$ is $T_{1}$ and time period of a uniform rod of the same length $l$ pivoted about one end and oscillating in a vertical plane is $T_{2}$. Amplitude of oscillations in both the cases is small. The, $T_{1} / T_{2}$ is
1 $\frac{1}{\sqrt{3}}$
2 1
3 $\sqrt{\frac{4}{3}}$
4 $\sqrt{\frac{3}{2}}$
Explanation:
D The period of simple pendulum is given by $\mathrm{T}_{1}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ Time period of uniform rod at given position - $\mathrm{T}_{2}=2 \pi \sqrt{\frac{\text { Inertia factor }}{\text { Spring factor }}}$ $\because \quad$ Inertia factor $=$ Moment of inertia of rod at one end $=\frac{\mathrm{m} l^{2}}{12}+\frac{\mathrm{m} l^{2}}{4}$ $=\frac{\mathrm{m} l^{2}+3 \mathrm{~m} l^{2}}{12}=\frac{4 \mathrm{~m} l^{2}}{12}=\frac{\mathrm{m} l^{2}}{3}$ And spring factor $=$ Restoring torque per unit angular displacement $=\mathrm{mg} \times \frac{l}{2} \frac{\sin \theta}{\theta}=\mathrm{mg} \frac{l}{2} \quad[\because \theta \text { is very small }]$ $\therefore \quad \mathrm{T}_{2} =2 \pi \sqrt{\frac{\mathrm{m} l^{2} / 3}{\mathrm{mgl} / 2}}$ $\mathrm{~T}_{2} =2 \pi \sqrt{\frac{2 l}{3 \mathrm{~g}}}$ $\text { Hence, } \frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}} =\sqrt{\frac{3}{2}}$
UP CPMT-2006
Oscillations
140579
The angular amplitude of simple pendulum is $\boldsymbol{\theta}_{0}$. The maximum tension in its string will be
D Let, mass of bob $=\mathrm{m}$, string length $=l$ Maximum tension in the string $\left(\mathrm{T}_{\text {max. }}\right)=m g+\frac{m v^{2}}{l}$ $=\mathrm{mg}+\frac{2 \mathrm{mg} l}{l}\left(1-\cos \theta_{0}\right)$ \(=\mathrm{mg}+\frac{2 \mathrm{mg} l}{l}\left(1-\cos \theta_0\right) \quad\left[\begin{array}{rl}\because \cos \theta_0 & =\frac{l-\mathrm{h}}{l} \\ \mathrm{v}^2 & =2 \mathrm{gh}\end{array}\right]\) $=\mathrm{mg}+2 \mathrm{mg}\left(1-1+\frac{\theta_{0}^{2}}{2}\right) \quad\left(\theta_{0} \text { is small }\right)$ $=\operatorname{mg}\left(1+\theta_{0}^{2}\right)$
140574
Two pendulum oscillate with a constant phase difference of $45^{\circ}$ and same amplitude. If the maximum velocity of one of them is $v$ and that of other is $v+x$, then the value of $x$ will be
1 0
2 $\frac{\mathrm{V}}{2}$
3 $\frac{\mathrm{V}}{\sqrt{2}}$
4 $(\sqrt{2}) \mathrm{v}$
Explanation:
A Let, velocity of one pendulum is $\mathrm{v}_{1}=\mathrm{v}_{\mathrm{o}} \cos \left(\omega \mathrm{t}+\phi_{1}\right)$ Velocity of the other pendulum is- $\mathrm{v}_{2}=\mathrm{v}_{\mathrm{o}} \cos \left(\omega \mathrm{t}+\phi_{2}\right)$ According to the question- $\phi_{2}-\phi_{1}=45^{\circ}$ Clearly, $\left|\mathrm{v}_{1}\right|_{\max }=\left|\mathrm{v}_{2}\right|_{\max }=\mathrm{v}_{\mathrm{o}}=\mathrm{v}$ Thus, $\quad \mathrm{x}=0$
J and K-CET-2015
Oscillations
140576
The period of oscillation of a simple pendulum of length $l$ suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination $\alpha$, is given by
A Given, simple pendulum of length $=l$ Acceleration $=g \sin \theta$. Since vehicle is accelerating a pseudo force $\mathrm{mg} \sin \theta$ will act on bob of pendulum which cancel the $\sin \theta$ component of weight of the bob. Hence, we can say that the effective acceleration would $=\mathrm{g} \cos \alpha$ Now, the time period of oscillation- $\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g} \cos \alpha}}$
JEE Main-29.07.2022
Oscillations
140578
Time period of a simple pendulum of length $l$ is $T_{1}$ and time period of a uniform rod of the same length $l$ pivoted about one end and oscillating in a vertical plane is $T_{2}$. Amplitude of oscillations in both the cases is small. The, $T_{1} / T_{2}$ is
1 $\frac{1}{\sqrt{3}}$
2 1
3 $\sqrt{\frac{4}{3}}$
4 $\sqrt{\frac{3}{2}}$
Explanation:
D The period of simple pendulum is given by $\mathrm{T}_{1}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ Time period of uniform rod at given position - $\mathrm{T}_{2}=2 \pi \sqrt{\frac{\text { Inertia factor }}{\text { Spring factor }}}$ $\because \quad$ Inertia factor $=$ Moment of inertia of rod at one end $=\frac{\mathrm{m} l^{2}}{12}+\frac{\mathrm{m} l^{2}}{4}$ $=\frac{\mathrm{m} l^{2}+3 \mathrm{~m} l^{2}}{12}=\frac{4 \mathrm{~m} l^{2}}{12}=\frac{\mathrm{m} l^{2}}{3}$ And spring factor $=$ Restoring torque per unit angular displacement $=\mathrm{mg} \times \frac{l}{2} \frac{\sin \theta}{\theta}=\mathrm{mg} \frac{l}{2} \quad[\because \theta \text { is very small }]$ $\therefore \quad \mathrm{T}_{2} =2 \pi \sqrt{\frac{\mathrm{m} l^{2} / 3}{\mathrm{mgl} / 2}}$ $\mathrm{~T}_{2} =2 \pi \sqrt{\frac{2 l}{3 \mathrm{~g}}}$ $\text { Hence, } \frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}} =\sqrt{\frac{3}{2}}$
UP CPMT-2006
Oscillations
140579
The angular amplitude of simple pendulum is $\boldsymbol{\theta}_{0}$. The maximum tension in its string will be
D Let, mass of bob $=\mathrm{m}$, string length $=l$ Maximum tension in the string $\left(\mathrm{T}_{\text {max. }}\right)=m g+\frac{m v^{2}}{l}$ $=\mathrm{mg}+\frac{2 \mathrm{mg} l}{l}\left(1-\cos \theta_{0}\right)$ \(=\mathrm{mg}+\frac{2 \mathrm{mg} l}{l}\left(1-\cos \theta_0\right) \quad\left[\begin{array}{rl}\because \cos \theta_0 & =\frac{l-\mathrm{h}}{l} \\ \mathrm{v}^2 & =2 \mathrm{gh}\end{array}\right]\) $=\mathrm{mg}+2 \mathrm{mg}\left(1-1+\frac{\theta_{0}^{2}}{2}\right) \quad\left(\theta_{0} \text { is small }\right)$ $=\operatorname{mg}\left(1+\theta_{0}^{2}\right)$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Oscillations
140574
Two pendulum oscillate with a constant phase difference of $45^{\circ}$ and same amplitude. If the maximum velocity of one of them is $v$ and that of other is $v+x$, then the value of $x$ will be
1 0
2 $\frac{\mathrm{V}}{2}$
3 $\frac{\mathrm{V}}{\sqrt{2}}$
4 $(\sqrt{2}) \mathrm{v}$
Explanation:
A Let, velocity of one pendulum is $\mathrm{v}_{1}=\mathrm{v}_{\mathrm{o}} \cos \left(\omega \mathrm{t}+\phi_{1}\right)$ Velocity of the other pendulum is- $\mathrm{v}_{2}=\mathrm{v}_{\mathrm{o}} \cos \left(\omega \mathrm{t}+\phi_{2}\right)$ According to the question- $\phi_{2}-\phi_{1}=45^{\circ}$ Clearly, $\left|\mathrm{v}_{1}\right|_{\max }=\left|\mathrm{v}_{2}\right|_{\max }=\mathrm{v}_{\mathrm{o}}=\mathrm{v}$ Thus, $\quad \mathrm{x}=0$
J and K-CET-2015
Oscillations
140576
The period of oscillation of a simple pendulum of length $l$ suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination $\alpha$, is given by
A Given, simple pendulum of length $=l$ Acceleration $=g \sin \theta$. Since vehicle is accelerating a pseudo force $\mathrm{mg} \sin \theta$ will act on bob of pendulum which cancel the $\sin \theta$ component of weight of the bob. Hence, we can say that the effective acceleration would $=\mathrm{g} \cos \alpha$ Now, the time period of oscillation- $\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g} \cos \alpha}}$
JEE Main-29.07.2022
Oscillations
140578
Time period of a simple pendulum of length $l$ is $T_{1}$ and time period of a uniform rod of the same length $l$ pivoted about one end and oscillating in a vertical plane is $T_{2}$. Amplitude of oscillations in both the cases is small. The, $T_{1} / T_{2}$ is
1 $\frac{1}{\sqrt{3}}$
2 1
3 $\sqrt{\frac{4}{3}}$
4 $\sqrt{\frac{3}{2}}$
Explanation:
D The period of simple pendulum is given by $\mathrm{T}_{1}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}$ Time period of uniform rod at given position - $\mathrm{T}_{2}=2 \pi \sqrt{\frac{\text { Inertia factor }}{\text { Spring factor }}}$ $\because \quad$ Inertia factor $=$ Moment of inertia of rod at one end $=\frac{\mathrm{m} l^{2}}{12}+\frac{\mathrm{m} l^{2}}{4}$ $=\frac{\mathrm{m} l^{2}+3 \mathrm{~m} l^{2}}{12}=\frac{4 \mathrm{~m} l^{2}}{12}=\frac{\mathrm{m} l^{2}}{3}$ And spring factor $=$ Restoring torque per unit angular displacement $=\mathrm{mg} \times \frac{l}{2} \frac{\sin \theta}{\theta}=\mathrm{mg} \frac{l}{2} \quad[\because \theta \text { is very small }]$ $\therefore \quad \mathrm{T}_{2} =2 \pi \sqrt{\frac{\mathrm{m} l^{2} / 3}{\mathrm{mgl} / 2}}$ $\mathrm{~T}_{2} =2 \pi \sqrt{\frac{2 l}{3 \mathrm{~g}}}$ $\text { Hence, } \frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}} =\sqrt{\frac{3}{2}}$
UP CPMT-2006
Oscillations
140579
The angular amplitude of simple pendulum is $\boldsymbol{\theta}_{0}$. The maximum tension in its string will be
D Let, mass of bob $=\mathrm{m}$, string length $=l$ Maximum tension in the string $\left(\mathrm{T}_{\text {max. }}\right)=m g+\frac{m v^{2}}{l}$ $=\mathrm{mg}+\frac{2 \mathrm{mg} l}{l}\left(1-\cos \theta_{0}\right)$ \(=\mathrm{mg}+\frac{2 \mathrm{mg} l}{l}\left(1-\cos \theta_0\right) \quad\left[\begin{array}{rl}\because \cos \theta_0 & =\frac{l-\mathrm{h}}{l} \\ \mathrm{v}^2 & =2 \mathrm{gh}\end{array}\right]\) $=\mathrm{mg}+2 \mathrm{mg}\left(1-1+\frac{\theta_{0}^{2}}{2}\right) \quad\left(\theta_{0} \text { is small }\right)$ $=\operatorname{mg}\left(1+\theta_{0}^{2}\right)$
