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**Q 1. Average velocity of a particle executing SHM in one complete vibration is: [2019]A: Aω/2B: AωC: D: ZeroAns: **D

In one complete vibration, displacement is zero. So, average velocity in one complete vibration

Then the amplitude of its oscillation is given by : [2019]

A:

B:

C:

D: A + B

Ans:

y = A

Equate SHM

y' = y – A

Resultant amplitude

y - projection of the radius vector of rotating particle P is: [2019]

A: y(t) = –3 cos2πt, where y in m

B:

C:

D:

Ans:

Solution:

At t = 0, y displacement is maximum, so equation will be cosine function.

T = 4s

y = a cosωt**Q 4. A pendulum is hung from the roof of a sufficiently high building and is moving freely to and fro like a simple harmonic oscillator. The acceleration of the bob of the pendulum is 20 m/s ^{2} at a distance of 5 m from the mean position. The time period of oscillation is:- [2018]A: 2πsB: πsC: 2sD: 1sAns:** B

A:

B:

C:

D:

Ans:

**Q 6. A particle is executing SHM along a straight line. Its velocities at distances x _{1} and x_{2} from the mean position are V_{1} and V_{2}, respectively. Its time period is: [2015]**

A:

B:

C:

D:

Solution:

A: simple harmonic with amplitude (a+b)/2

B: not a simple harmonic

C: simple harmonic with amplitude a/b

D: simple harmonic with amplitude

Ans:

where c

The superimposed motion is simple harmonic with amplitude

X = A cos (ωt)

Where X = displacement at time t

ω = frequency of oscillation

Which one of the following graphs shows correctly the variation ‘a’ with ‘t’? [2014]

A:

B:

C:

D:

**Here, a = acceleration at time ‘t’ and T = time periodAns:** A

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