NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI14:OSCILLATIONS
364279
Average velocity of a particle executing SHM in one complete vibration is :
1 \(\dfrac{A \omega}{2}\)
2 \(A \omega\)
3 \(\dfrac{A \omega^{2}}{2}\)
4 Zero
Explanation:
In one complete vibration, displacement is zero. So, average velocity in one complete vibration \( = \frac{{{\rm{ Displacement }}}}{{{\rm{ Time}}\,\,{\rm{interval }}}} = 0\)
NEET - 2019
PHXI14:OSCILLATIONS
364280
A \(1.00 \times {10^{ - 20}}\;kg\) particle is executing with simple harmonic motion with a period of \(1.00 \times {10^{ - 5}}\sec \) and a maximum speed of \(1.00 \times {10^3}\;m{\rm{/}}s\). The maximum displacement of the particle is:
364281
If displacement \(x\) and velocity \(v\) are related as \(4 v^{2}=25-x^{2}\) in a SHM then time period of given SHM is (consider SI units)
1 \(2 \pi\)
2 \(\pi\)
3 \(6 \pi\)
4 \(4 \pi\)
Explanation:
Given \(4 v^{2}=25-x^{2}\) Differentiating with respect to \(t\) on both sides, \(8 v a=-2 x v \Rightarrow a=-\dfrac{x}{4}\) Motion is simple harmonic: So, \(\omega^{2}=\dfrac{1}{4}\) and \(T=2 \pi \sqrt{4}=4 \pi\)
PHXI14:OSCILLATIONS
364282
For a particle executing simple harmonic motion (SHM), at its mean position
1 Velocity is zero and acceleration is maximum
2 Velocity is maximum and acceleration is zero
3 Both velocity and acceleration are maximum
4 Both velocity and acceleration are zero
Explanation:
The relation between velocity, acceleration and displacement is given by \({V=\omega \sqrt{A^{2}-y^{2}}}\) \({a=-\omega^{2} y}\) At mean position \({y=0}\) \({\therefore V=}\) maximum and \({a=}\) minimum.
364279
Average velocity of a particle executing SHM in one complete vibration is :
1 \(\dfrac{A \omega}{2}\)
2 \(A \omega\)
3 \(\dfrac{A \omega^{2}}{2}\)
4 Zero
Explanation:
In one complete vibration, displacement is zero. So, average velocity in one complete vibration \( = \frac{{{\rm{ Displacement }}}}{{{\rm{ Time}}\,\,{\rm{interval }}}} = 0\)
NEET - 2019
PHXI14:OSCILLATIONS
364280
A \(1.00 \times {10^{ - 20}}\;kg\) particle is executing with simple harmonic motion with a period of \(1.00 \times {10^{ - 5}}\sec \) and a maximum speed of \(1.00 \times {10^3}\;m{\rm{/}}s\). The maximum displacement of the particle is:
364281
If displacement \(x\) and velocity \(v\) are related as \(4 v^{2}=25-x^{2}\) in a SHM then time period of given SHM is (consider SI units)
1 \(2 \pi\)
2 \(\pi\)
3 \(6 \pi\)
4 \(4 \pi\)
Explanation:
Given \(4 v^{2}=25-x^{2}\) Differentiating with respect to \(t\) on both sides, \(8 v a=-2 x v \Rightarrow a=-\dfrac{x}{4}\) Motion is simple harmonic: So, \(\omega^{2}=\dfrac{1}{4}\) and \(T=2 \pi \sqrt{4}=4 \pi\)
PHXI14:OSCILLATIONS
364282
For a particle executing simple harmonic motion (SHM), at its mean position
1 Velocity is zero and acceleration is maximum
2 Velocity is maximum and acceleration is zero
3 Both velocity and acceleration are maximum
4 Both velocity and acceleration are zero
Explanation:
The relation between velocity, acceleration and displacement is given by \({V=\omega \sqrt{A^{2}-y^{2}}}\) \({a=-\omega^{2} y}\) At mean position \({y=0}\) \({\therefore V=}\) maximum and \({a=}\) minimum.
364279
Average velocity of a particle executing SHM in one complete vibration is :
1 \(\dfrac{A \omega}{2}\)
2 \(A \omega\)
3 \(\dfrac{A \omega^{2}}{2}\)
4 Zero
Explanation:
In one complete vibration, displacement is zero. So, average velocity in one complete vibration \( = \frac{{{\rm{ Displacement }}}}{{{\rm{ Time}}\,\,{\rm{interval }}}} = 0\)
NEET - 2019
PHXI14:OSCILLATIONS
364280
A \(1.00 \times {10^{ - 20}}\;kg\) particle is executing with simple harmonic motion with a period of \(1.00 \times {10^{ - 5}}\sec \) and a maximum speed of \(1.00 \times {10^3}\;m{\rm{/}}s\). The maximum displacement of the particle is:
364281
If displacement \(x\) and velocity \(v\) are related as \(4 v^{2}=25-x^{2}\) in a SHM then time period of given SHM is (consider SI units)
1 \(2 \pi\)
2 \(\pi\)
3 \(6 \pi\)
4 \(4 \pi\)
Explanation:
Given \(4 v^{2}=25-x^{2}\) Differentiating with respect to \(t\) on both sides, \(8 v a=-2 x v \Rightarrow a=-\dfrac{x}{4}\) Motion is simple harmonic: So, \(\omega^{2}=\dfrac{1}{4}\) and \(T=2 \pi \sqrt{4}=4 \pi\)
PHXI14:OSCILLATIONS
364282
For a particle executing simple harmonic motion (SHM), at its mean position
1 Velocity is zero and acceleration is maximum
2 Velocity is maximum and acceleration is zero
3 Both velocity and acceleration are maximum
4 Both velocity and acceleration are zero
Explanation:
The relation between velocity, acceleration and displacement is given by \({V=\omega \sqrt{A^{2}-y^{2}}}\) \({a=-\omega^{2} y}\) At mean position \({y=0}\) \({\therefore V=}\) maximum and \({a=}\) minimum.
364279
Average velocity of a particle executing SHM in one complete vibration is :
1 \(\dfrac{A \omega}{2}\)
2 \(A \omega\)
3 \(\dfrac{A \omega^{2}}{2}\)
4 Zero
Explanation:
In one complete vibration, displacement is zero. So, average velocity in one complete vibration \( = \frac{{{\rm{ Displacement }}}}{{{\rm{ Time}}\,\,{\rm{interval }}}} = 0\)
NEET - 2019
PHXI14:OSCILLATIONS
364280
A \(1.00 \times {10^{ - 20}}\;kg\) particle is executing with simple harmonic motion with a period of \(1.00 \times {10^{ - 5}}\sec \) and a maximum speed of \(1.00 \times {10^3}\;m{\rm{/}}s\). The maximum displacement of the particle is:
364281
If displacement \(x\) and velocity \(v\) are related as \(4 v^{2}=25-x^{2}\) in a SHM then time period of given SHM is (consider SI units)
1 \(2 \pi\)
2 \(\pi\)
3 \(6 \pi\)
4 \(4 \pi\)
Explanation:
Given \(4 v^{2}=25-x^{2}\) Differentiating with respect to \(t\) on both sides, \(8 v a=-2 x v \Rightarrow a=-\dfrac{x}{4}\) Motion is simple harmonic: So, \(\omega^{2}=\dfrac{1}{4}\) and \(T=2 \pi \sqrt{4}=4 \pi\)
PHXI14:OSCILLATIONS
364282
For a particle executing simple harmonic motion (SHM), at its mean position
1 Velocity is zero and acceleration is maximum
2 Velocity is maximum and acceleration is zero
3 Both velocity and acceleration are maximum
4 Both velocity and acceleration are zero
Explanation:
The relation between velocity, acceleration and displacement is given by \({V=\omega \sqrt{A^{2}-y^{2}}}\) \({a=-\omega^{2} y}\) At mean position \({y=0}\) \({\therefore V=}\) maximum and \({a=}\) minimum.
