Q1: If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length, then the new time period of oscillation is x / 2 times its original time period. Then the value of x is: [2024]
(a) √3
(b) √2
(c) √2
(d) 4
Ans: (b)
The period of oscillation, T, of a simple pendulum is determined by the formula:
where:
The mass of the bob does not factor into the equation for the period.
Let's first denote the original length of the pendulum as L and the original period of oscillation as T_{1}. Hence,
When the length of the pendulum is halved, the new length L' would be L / 2. Thus, the new period T_{2} can be calculated as:
We are given that the new period T_{2}
To find the value of x, we solve for x:
Multiplying both sides by 2:
Simplify to:
Hence, the correct answer is:
Option B:
(a) 5 cm, 2 s
(b) 5 m, 2 s
(c) 5 cm, 1 s
(d) 5 m, 1 s
Ans: (b)
In the equation for simple harmonic motion (SHM), , the general form can be used to identify the parameters of SHM, where:
Comparing the given equation with the standard form:
The angular frequency ω is related to the time period T of the motion through the formula:
Given that ω
Hence, the amplitude of the motion is 5 m and the time period is 2 s. Thus, the correct answer is:
Option B: 5 m, 2 s
Q1: The xt graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at t = 2 s is [2023]
(a)
(b)
(c)
(d)
Ans: (d)
Solution: Position of particle as function of time
Q1: Match ListI with ListII [2022]
Choose the correct answer from the options given below
(a) (a)  (iii), (b)  (ii), (c)  (i), (d)  (iv)
(b) (a)  (iv), (b)  (ii), (c)  (iii), (d)  (i)
(c) (a)  (iv), (b)  (iii), (c)  (ii), (d)  (i)
(d) (a)  (i), (b)  (iv), (c)  (iii), (d)  (ii)
Ans: (c)
(a) Amplitude of oscillation is continuously decreasing. It means bob of pendulum oscillate with air friction.
(b) F = kx (restoring force of a spring)
(c) Amplitude of oscillation is remains same. It means bob of pendulum is oscillating under negligible air resistance.
(d)
Q2: Identify the function which represents a nonperiodic motion. [2022]
(a)
(b) e^{ωt}
(c) sinωt
(d) sinωt + cosωt
Ans: (b)
Here,
And
While,
Q3: Two pendulums of length 121 cm and 100 cm start vibrating in phase. At some instant, the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is: [2022]
(a) 9
(b) 10
(c) 8
(d) 11
Ans: (d)
Solution:
Q1: A body is executing simple harmonic motion with frequency 'n', the frequency of its potential energy is: [2021]
(a) 3n
(b) 4n
(c) n
(d) 2n
Ans: (d)
Solution: Displacement equation of SHM of frequency 'n' x = Asin(ωt) = Asin(2πnt) Now,
So frequency of potential energy = 2n
Q2: A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is: [2021]
(a) 3.14 s
(b) 0.628 s
(c) 0.0628 s
(d) 6.28 s
Ans:
Solution:
F = kx
10 = k(5 × 10^{–2})
= 2 x 10^{2}
= 200 N/m
Now,
Q1: The phase difference between displacement and acceleration of a particle in a simple harmonic motion is :
(a) 3π/2 rad
(b) π/2 rad
(c) Zero
(d) π rad [2020]
Ans: (d)
Q1: Average velocity of a particle executing SHM in one complete vibration is: [2019]
(a) Aω/2
(b) Aω
(c)
(d) Zero
Ans: (d)
Solution:
In one complete vibration, displacement is zero. So, average velocity in one complete vibration
Q2: The displacement of a particle executing simple harmonic motion is given by y = A_{0}+ Asinωt + Bcosωt
Then the amplitude of its oscillation is given by : [2019]
(a)
(b)
(c)
(d) A + B
Ans: (b)
Solution:
y = A_{0}+ Asinωt + Bsinωt
Equate SHM
y' = y – A_{0} = Asinωt + Bcosωt
Resultant amplitude
Q3: The radius of circle, the period of revolution, initial position and sense of revolution are indicated in the fig.
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: (d)
Solution:
At t = 0, y displacement is maximum, so equation will be cosine function.
T = 4s
y = a cosωt
Q1: 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) 2πs
(b) πs
(c) 2s
(d) 1s
Ans: (b)
Solution:
Q1: A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is: [2017]
(a)
(b)
(c)
(d)
Ans: (b)
Solution:
Let us assume, the length of spring be l.
When we cut the spring into ratio of length 1 : 2 : 3, we get three springs of lengths with force constant,
When connected in series,
When connected in parallel,
On dividing :
Q1: 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: [NEET /AIPMT Cancelled Paper 2015]
(a)
(b)
(c)
(d)
Ans: (c)
Solution:
Q2: When two displacements represented by y_{1} = a sin(ωt) and y_{2} = b cos(ωt) are superimposed the motion is: [NEET /AIPMT Cancelled Paper 2015]
(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: (d)
Solution:
where c^{2} = a^{2} + b^{2 }[since a^{2 }+ b^{2} = c^{2} cos (φ) + c^{2 }sin (φ) = c^{2}^{ }]
The superimposed motion is simple harmonic with amplitude
Q3: A particle is executing a simple harmonic motion. Its maximum acceleration is α and maximum velocity is β. Them, its time period of vibration will be [NEET / AIPMT 2015]
(a) β^{2}/ α
(b) 2πβ / α
(c) β^{2}/ α^{2}
(d) α / β
Ans: (b)
As, we know, in SHM
Q1: The oscillation of a body on a smooth horizontal surface is represented by the equation,
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 period
Ans: (a)
Solution:
102 videos411 docs121 tests

1. What is the formula for the time period of an oscillating springmass system? 
2. How does the amplitude of oscillation affect the energy of a system? 
3. What is the relationship between the frequency and the time period of an oscillating system? 
4. How does damping affect the motion of an oscillating system? 
5. What is the difference between free oscillations and forced oscillations? 

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