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NEET Exam  >  NEET Tests  >  Test: Osillations (October 15) - NEET MCQ

Test: Osillations (October 15) - NEET MCQ


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10 Questions MCQ Test - Test: Osillations (October 15)

Test: Osillations (October 15) for NEET 2024 is part of NEET preparation. The Test: Osillations (October 15) questions and answers have been prepared according to the NEET exam syllabus.The Test: Osillations (October 15) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Osillations (October 15) below.
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Test: Osillations (October 15) - Question 1
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The pendulum of a wall clock executes

Detailed Solution for Test: Osillations (October 15) - Question 1

The bob of a pendulum moves in such a way that it repeats its positions several time but alternately the time gap between these positions is always equal which proves that the motion of a pendulum is oscillatory.

Test: Osillations (October 15) - Question 2
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The kinetic energy of a body executing S.H.M. is 1/3 of the potential energy. Then, the displacement of the body is x percent of the amplitude, where x is

Detailed Solution for Test: Osillations (October 15) - Question 2

We know that PE + KE = TE = constant
Hence at the extremis TE = PE = ½ kz2  
Where z is amplitude and k is shm constant.
Thus when KE = ⅓ PE
We get PE = ¾ TE =  ½ kz2
Hence we get  ½ kx2 = ¾  ½ kz2
We get x/z = √3/4 
= 1.73 / 2
= .87
Thus we get x is 87 percent of the amplitude.

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Test: Osillations (October 15) - Question 3
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A body is executing S.H.M When its displacement from the mean position is 4 cm and 5 cm, the corresponding velocities of the body are 10 cm per sec and 8 cm per sec. Then the time period of the body is

Detailed Solution for Test: Osillations (October 15) - Question 3

v=ω√A2−x2
​⇒8=ω √A2−25​
&10=ω√A2−16
​⇒(8/10​)2= A2−25/ A2−16
​⇒16/25​= A2−25/ A2−16
​⇒16A2−256=25A2−625
⇒9A2=369
A2=41
⇒8=ω41−25
​⇒8=4ω
∴ω=2=T2π​
T=π sec​

Test: Osillations (October 15) - Question 4
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In a damped system
Test: Osillations (October 15) - Question 5
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Two pendulums have time periods T and 5T/4. They are in phase at their mean positions at some instant of time. What will be their phase difference when the bigger pendulum completes one oscillation?
Detailed Solution for Test: Osillations (October 15) - Question 5

https://cn.edurev.in/ApplicationImages/Temp/540839_ead174e5-cd42-4612-ab3a-d5f431628c54_lg.png

Test: Osillations (October 15) - Question 6
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Periodic motion is
Test: Osillations (October 15) - Question 7
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Two pendulums of length 1 m and 16 m start vibrating one behind the other from the same stand. At some instant, the two are in the mean position in the same phase. The time period of shorter pendulum is T. The minimum time after which the two threads of the pendulum will be one behind the other is
Detailed Solution for Test: Osillations (October 15) - Question 7

T1=2π√1/g=T, T2=2π√16/g=4T
(ω1−ω2)t=2π
(2π/T−2π/4T)t=2π
t=4T/3

Test: Osillations (October 15) - Question 8
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A particle is executing S.H.M with an amplitude of 4 cm and time period 12 sec. The time taken by the particle in going from its mean position to a position of displacement equal to 2 cm is T1..The time taken from this displaced position of 2 cm to reach the extreme position is T2. Therefore, T1/ T2 will be
Test: Osillations (October 15) - Question 9
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In simple models damping force is
Test: Osillations (October 15) - Question 10
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Whose motion is an ideal case of SHM?
Detailed Solution for Test: Osillations (October 15) - Question 10

Simple pendulum is not a case of ideal SHM as it is mention in NCERT.
Shadow of a point mass having uniform circular motion on the horizontal diameter is a case of SHM but not on the vertical diameter.

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