NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI14:OSCILLATIONS
364168
A particle is executing SHM of periodic time \(T\). The time taken by a particle in moving from mean position to half the maximum displacement is \(\left( {\sin {{30}^o} = 0.5} \right)\)
1 \(\frac{T}{2}\)
2 \(\frac{T}{4}\)
3 \(\frac{T}{8}\)
4 \(\frac{T}{{12}}\)
Explanation:
If the particle at \(t = 0\;s\) is at mean position, the maximum displacement is given by \(x=A \sin \dfrac{2 \pi t}{T}\) If it takes time \(t_{1}\) to move from mean position to half the displacemnt, then \(\frac{A}{2} = A\sin \frac{{2\pi t}}{T}{\rm{ or }}\frac{1}{2} = \sin \frac{{2\pi t}}{T}{\rm{ }}\) \({\rm{or }}\sin {30^{\rm{o}}} = \sin \frac{{2\pi t}}{T}\;\;\;1pt\sin \,{30^{\rm{o}}} = \frac{1}{2}\) \(\sin \frac{\pi }{6} = \sin \frac{{2\pi t}}{T}\) \(\frac{\pi }{6} = \frac{{2\pi t}}{T}\) \( \Rightarrow t = \frac{T}{{12}}s\)
MHTCET - 2015
PHXI14:OSCILLATIONS
364169
Two pendulums of different lengths are in phase at the mean position at a certain instant. The minimum time after which they will be again in phase is \(5\;T/4\), where \(T\) is the time period of shorter pendulum. Find the ratio of lengths of the two pendulums.
1 \(1: 4\)
2 \(1: 16\)
3 \(1: 25\)
4 \(1: 2\)
Explanation:
Let \(T_{1}=\mathrm{T}_{\text {and }} \mathrm{T}_{2}=K T ; \dfrac{2 \pi}{T} t-\dfrac{2 \pi}{K T} t=2 \pi\) \(t\left(\dfrac{K-1}{K T}\right)=1 ; \dfrac{5 T}{4}\left(\dfrac{K-1}{K T}\right)=1\) which gives \(K = 5\) If length of first pendulum \(=l, T_{1}=2 \pi \sqrt{l / g}\) \(T_{2}=5 \times 2 \pi \sqrt{l / g}=2 \pi \sqrt{25 l / g}\) So, ratio of lengths \(=1: 25\)
PHXI14:OSCILLATIONS
364170
A particle is performing simple haromonic motion along \(x\)-axis with amplitude \(4\;cm\) and time period \(1.2\,\sec \). The minimum time taken by the particle to move from \(x = 2\;cm\) to \(x = + 4\;cm\) and back again is given by [Mean position is at \(x=0\) ]
1 \(0.4\,{\rm{sec}}\)
2 \(0.6\,{\rm{sec}}\)
3 \(0.2\,{\rm{sec}}\)
4 \(0.3\,{\rm{sec}}\)
Explanation:
Time taken by particle to move from \(x=\) 0 (mean position) to \(x=4\) (extreme position) \( = \frac{T}{4} = \frac{{1.2}}{4} = 0.3\;s\) Let \(t\) be the time taken by the particle to move from \(x = 0{\rm{ }}to{\rm{ }}x = 2\;cm\) \(y = a\sin \omega t \Rightarrow 2 = 4\sin \frac{{2\pi }}{T}t \Rightarrow \frac{1}{2} = \sin \frac{{2\pi }}{{1.2}}t\) \(\Rightarrow \dfrac{\pi}{6}=\dfrac{2 \pi}{1.2} t \Rightarrow t=0.1 s\). Hence time to move from \(x=2\) to \(x=4\) will be equal to \(0.3-0.1=0.2 s\) Hence total time to move from \(x=2\) to \(x=4\) and back again \(=2 \times 0.2=0.4 \mathrm{sec}\)
PHXI14:OSCILLATIONS
364171
Assertion : Acceleration is proportional to the displacement. This condition is not sufficient for motion to be simple harmonic. Reason : In simple harmonic motion direction of displacement is also considered.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
In SHM, the acceleration is always in a direction opposite to that of the displacement i.e., proportional to \((-y)\).
364168
A particle is executing SHM of periodic time \(T\). The time taken by a particle in moving from mean position to half the maximum displacement is \(\left( {\sin {{30}^o} = 0.5} \right)\)
1 \(\frac{T}{2}\)
2 \(\frac{T}{4}\)
3 \(\frac{T}{8}\)
4 \(\frac{T}{{12}}\)
Explanation:
If the particle at \(t = 0\;s\) is at mean position, the maximum displacement is given by \(x=A \sin \dfrac{2 \pi t}{T}\) If it takes time \(t_{1}\) to move from mean position to half the displacemnt, then \(\frac{A}{2} = A\sin \frac{{2\pi t}}{T}{\rm{ or }}\frac{1}{2} = \sin \frac{{2\pi t}}{T}{\rm{ }}\) \({\rm{or }}\sin {30^{\rm{o}}} = \sin \frac{{2\pi t}}{T}\;\;\;1pt\sin \,{30^{\rm{o}}} = \frac{1}{2}\) \(\sin \frac{\pi }{6} = \sin \frac{{2\pi t}}{T}\) \(\frac{\pi }{6} = \frac{{2\pi t}}{T}\) \( \Rightarrow t = \frac{T}{{12}}s\)
MHTCET - 2015
PHXI14:OSCILLATIONS
364169
Two pendulums of different lengths are in phase at the mean position at a certain instant. The minimum time after which they will be again in phase is \(5\;T/4\), where \(T\) is the time period of shorter pendulum. Find the ratio of lengths of the two pendulums.
1 \(1: 4\)
2 \(1: 16\)
3 \(1: 25\)
4 \(1: 2\)
Explanation:
Let \(T_{1}=\mathrm{T}_{\text {and }} \mathrm{T}_{2}=K T ; \dfrac{2 \pi}{T} t-\dfrac{2 \pi}{K T} t=2 \pi\) \(t\left(\dfrac{K-1}{K T}\right)=1 ; \dfrac{5 T}{4}\left(\dfrac{K-1}{K T}\right)=1\) which gives \(K = 5\) If length of first pendulum \(=l, T_{1}=2 \pi \sqrt{l / g}\) \(T_{2}=5 \times 2 \pi \sqrt{l / g}=2 \pi \sqrt{25 l / g}\) So, ratio of lengths \(=1: 25\)
PHXI14:OSCILLATIONS
364170
A particle is performing simple haromonic motion along \(x\)-axis with amplitude \(4\;cm\) and time period \(1.2\,\sec \). The minimum time taken by the particle to move from \(x = 2\;cm\) to \(x = + 4\;cm\) and back again is given by [Mean position is at \(x=0\) ]
1 \(0.4\,{\rm{sec}}\)
2 \(0.6\,{\rm{sec}}\)
3 \(0.2\,{\rm{sec}}\)
4 \(0.3\,{\rm{sec}}\)
Explanation:
Time taken by particle to move from \(x=\) 0 (mean position) to \(x=4\) (extreme position) \( = \frac{T}{4} = \frac{{1.2}}{4} = 0.3\;s\) Let \(t\) be the time taken by the particle to move from \(x = 0{\rm{ }}to{\rm{ }}x = 2\;cm\) \(y = a\sin \omega t \Rightarrow 2 = 4\sin \frac{{2\pi }}{T}t \Rightarrow \frac{1}{2} = \sin \frac{{2\pi }}{{1.2}}t\) \(\Rightarrow \dfrac{\pi}{6}=\dfrac{2 \pi}{1.2} t \Rightarrow t=0.1 s\). Hence time to move from \(x=2\) to \(x=4\) will be equal to \(0.3-0.1=0.2 s\) Hence total time to move from \(x=2\) to \(x=4\) and back again \(=2 \times 0.2=0.4 \mathrm{sec}\)
PHXI14:OSCILLATIONS
364171
Assertion : Acceleration is proportional to the displacement. This condition is not sufficient for motion to be simple harmonic. Reason : In simple harmonic motion direction of displacement is also considered.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
In SHM, the acceleration is always in a direction opposite to that of the displacement i.e., proportional to \((-y)\).
364168
A particle is executing SHM of periodic time \(T\). The time taken by a particle in moving from mean position to half the maximum displacement is \(\left( {\sin {{30}^o} = 0.5} \right)\)
1 \(\frac{T}{2}\)
2 \(\frac{T}{4}\)
3 \(\frac{T}{8}\)
4 \(\frac{T}{{12}}\)
Explanation:
If the particle at \(t = 0\;s\) is at mean position, the maximum displacement is given by \(x=A \sin \dfrac{2 \pi t}{T}\) If it takes time \(t_{1}\) to move from mean position to half the displacemnt, then \(\frac{A}{2} = A\sin \frac{{2\pi t}}{T}{\rm{ or }}\frac{1}{2} = \sin \frac{{2\pi t}}{T}{\rm{ }}\) \({\rm{or }}\sin {30^{\rm{o}}} = \sin \frac{{2\pi t}}{T}\;\;\;1pt\sin \,{30^{\rm{o}}} = \frac{1}{2}\) \(\sin \frac{\pi }{6} = \sin \frac{{2\pi t}}{T}\) \(\frac{\pi }{6} = \frac{{2\pi t}}{T}\) \( \Rightarrow t = \frac{T}{{12}}s\)
MHTCET - 2015
PHXI14:OSCILLATIONS
364169
Two pendulums of different lengths are in phase at the mean position at a certain instant. The minimum time after which they will be again in phase is \(5\;T/4\), where \(T\) is the time period of shorter pendulum. Find the ratio of lengths of the two pendulums.
1 \(1: 4\)
2 \(1: 16\)
3 \(1: 25\)
4 \(1: 2\)
Explanation:
Let \(T_{1}=\mathrm{T}_{\text {and }} \mathrm{T}_{2}=K T ; \dfrac{2 \pi}{T} t-\dfrac{2 \pi}{K T} t=2 \pi\) \(t\left(\dfrac{K-1}{K T}\right)=1 ; \dfrac{5 T}{4}\left(\dfrac{K-1}{K T}\right)=1\) which gives \(K = 5\) If length of first pendulum \(=l, T_{1}=2 \pi \sqrt{l / g}\) \(T_{2}=5 \times 2 \pi \sqrt{l / g}=2 \pi \sqrt{25 l / g}\) So, ratio of lengths \(=1: 25\)
PHXI14:OSCILLATIONS
364170
A particle is performing simple haromonic motion along \(x\)-axis with amplitude \(4\;cm\) and time period \(1.2\,\sec \). The minimum time taken by the particle to move from \(x = 2\;cm\) to \(x = + 4\;cm\) and back again is given by [Mean position is at \(x=0\) ]
1 \(0.4\,{\rm{sec}}\)
2 \(0.6\,{\rm{sec}}\)
3 \(0.2\,{\rm{sec}}\)
4 \(0.3\,{\rm{sec}}\)
Explanation:
Time taken by particle to move from \(x=\) 0 (mean position) to \(x=4\) (extreme position) \( = \frac{T}{4} = \frac{{1.2}}{4} = 0.3\;s\) Let \(t\) be the time taken by the particle to move from \(x = 0{\rm{ }}to{\rm{ }}x = 2\;cm\) \(y = a\sin \omega t \Rightarrow 2 = 4\sin \frac{{2\pi }}{T}t \Rightarrow \frac{1}{2} = \sin \frac{{2\pi }}{{1.2}}t\) \(\Rightarrow \dfrac{\pi}{6}=\dfrac{2 \pi}{1.2} t \Rightarrow t=0.1 s\). Hence time to move from \(x=2\) to \(x=4\) will be equal to \(0.3-0.1=0.2 s\) Hence total time to move from \(x=2\) to \(x=4\) and back again \(=2 \times 0.2=0.4 \mathrm{sec}\)
PHXI14:OSCILLATIONS
364171
Assertion : Acceleration is proportional to the displacement. This condition is not sufficient for motion to be simple harmonic. Reason : In simple harmonic motion direction of displacement is also considered.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
In SHM, the acceleration is always in a direction opposite to that of the displacement i.e., proportional to \((-y)\).
364168
A particle is executing SHM of periodic time \(T\). The time taken by a particle in moving from mean position to half the maximum displacement is \(\left( {\sin {{30}^o} = 0.5} \right)\)
1 \(\frac{T}{2}\)
2 \(\frac{T}{4}\)
3 \(\frac{T}{8}\)
4 \(\frac{T}{{12}}\)
Explanation:
If the particle at \(t = 0\;s\) is at mean position, the maximum displacement is given by \(x=A \sin \dfrac{2 \pi t}{T}\) If it takes time \(t_{1}\) to move from mean position to half the displacemnt, then \(\frac{A}{2} = A\sin \frac{{2\pi t}}{T}{\rm{ or }}\frac{1}{2} = \sin \frac{{2\pi t}}{T}{\rm{ }}\) \({\rm{or }}\sin {30^{\rm{o}}} = \sin \frac{{2\pi t}}{T}\;\;\;1pt\sin \,{30^{\rm{o}}} = \frac{1}{2}\) \(\sin \frac{\pi }{6} = \sin \frac{{2\pi t}}{T}\) \(\frac{\pi }{6} = \frac{{2\pi t}}{T}\) \( \Rightarrow t = \frac{T}{{12}}s\)
MHTCET - 2015
PHXI14:OSCILLATIONS
364169
Two pendulums of different lengths are in phase at the mean position at a certain instant. The minimum time after which they will be again in phase is \(5\;T/4\), where \(T\) is the time period of shorter pendulum. Find the ratio of lengths of the two pendulums.
1 \(1: 4\)
2 \(1: 16\)
3 \(1: 25\)
4 \(1: 2\)
Explanation:
Let \(T_{1}=\mathrm{T}_{\text {and }} \mathrm{T}_{2}=K T ; \dfrac{2 \pi}{T} t-\dfrac{2 \pi}{K T} t=2 \pi\) \(t\left(\dfrac{K-1}{K T}\right)=1 ; \dfrac{5 T}{4}\left(\dfrac{K-1}{K T}\right)=1\) which gives \(K = 5\) If length of first pendulum \(=l, T_{1}=2 \pi \sqrt{l / g}\) \(T_{2}=5 \times 2 \pi \sqrt{l / g}=2 \pi \sqrt{25 l / g}\) So, ratio of lengths \(=1: 25\)
PHXI14:OSCILLATIONS
364170
A particle is performing simple haromonic motion along \(x\)-axis with amplitude \(4\;cm\) and time period \(1.2\,\sec \). The minimum time taken by the particle to move from \(x = 2\;cm\) to \(x = + 4\;cm\) and back again is given by [Mean position is at \(x=0\) ]
1 \(0.4\,{\rm{sec}}\)
2 \(0.6\,{\rm{sec}}\)
3 \(0.2\,{\rm{sec}}\)
4 \(0.3\,{\rm{sec}}\)
Explanation:
Time taken by particle to move from \(x=\) 0 (mean position) to \(x=4\) (extreme position) \( = \frac{T}{4} = \frac{{1.2}}{4} = 0.3\;s\) Let \(t\) be the time taken by the particle to move from \(x = 0{\rm{ }}to{\rm{ }}x = 2\;cm\) \(y = a\sin \omega t \Rightarrow 2 = 4\sin \frac{{2\pi }}{T}t \Rightarrow \frac{1}{2} = \sin \frac{{2\pi }}{{1.2}}t\) \(\Rightarrow \dfrac{\pi}{6}=\dfrac{2 \pi}{1.2} t \Rightarrow t=0.1 s\). Hence time to move from \(x=2\) to \(x=4\) will be equal to \(0.3-0.1=0.2 s\) Hence total time to move from \(x=2\) to \(x=4\) and back again \(=2 \times 0.2=0.4 \mathrm{sec}\)
PHXI14:OSCILLATIONS
364171
Assertion : Acceleration is proportional to the displacement. This condition is not sufficient for motion to be simple harmonic. Reason : In simple harmonic motion direction of displacement is also considered.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
In SHM, the acceleration is always in a direction opposite to that of the displacement i.e., proportional to \((-y)\).
