Test: Osillations (October 15) - NEET MCQ

# Test: Osillations (October 15) - NEET MCQ

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## 10 Questions MCQ Test Daily Test for NEET Preparation - Test: Osillations (October 15)

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Test: Osillations (October 15) - Question 1

### 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

### 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

### 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

In a damped system

Test: Osillations (October 15) - Question 5

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

Test: Osillations (October 15) - Question 6

Periodic motion is

Test: Osillations (October 15) - Question 7

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

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

In simple models damping force is

Test: Osillations (October 15) - Question 10

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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