Simple Harmonic Motion NAT Level - 2
10 Questions MCQ Test Topic wise Tests for IIT JAM Physics | Simple Harmonic Motion NAT Level - 2
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A particle of mass m moves in one dimensional potential v(x)=-ax2 + bx4 where a and b are positive constants. What is the angular frequency of small oscillations about the minimum of the potential in units of 
A spring of force constant k is stretched a certain distance. It takes twice as much work to stretch a second spring by half this distance. What is the force constant of the second spring? (in units of k)
The correct answer is: 8
Figure shows block 1 of mass 0.2 g sliding to the right over a frictionless elevated surface at a speed of 8 m/s. The block undergoes an elastic collision with stationary block 2, which is attached to a spring of constant 1200 N/m. After the collision, block 2 oscillates in SHM with a period of 0.2 s & block 1 slides off the opposite end of the elevated surface, surface falling height = 6m find the value of d in meters
Time period of a SHM, 
Now, we can find the rebound speed of the block 1 after collision,
Negative sign indicates that the block 1 moves towards left
Time of flight = 
d = velocity x time
= 5.7 x 1 = 5.7 m
The correct answer is: 5.7
What is the period of oscillation of a physical pendulum which consists of a circular hoop hanging from a nail an a wall the mass of the hoop is 3 kg and its radius is 20 cm, if it displaced slightly by a passing breeze? (in seconds)
Parallel axis theorem
d → distance from pivot to centre of mass.
Period of physical pendulum
= 1.2 seconds
The correct answer is: 1.2
A long straight and massless rod pivots about one end in a vertical plane. In configuration I, two small identical masses are attached to the free end, in the second configuration one mass is attached at the centre and the other is attached to the free end. What is the ratio of frequency of small oscillations of configuration / to that of configuration II?
I = moment of inertia
r = moment arm
(for small angles)
∴ 
For the first system
l1 = 2mr2
radius of gyration = r1, = 2r
For the second system
radius of gyration 
The correct answer is: 1.09
Two identical springs with spring constant k are connected to identical masses of mass in, as shown in the figures below the ratio of the period for the springs connected in parallel (figure I) to the period for the springs connected in series (figure 2) is what?
Parallel
Effective spring constant = k1+ k2 = kp
Series
Effective spring constant 
The correct answer is: 0.5
What is the phase constant in degrees for the harmonic oscillator with the velocity function V(f) given in figure if the position function x(t) has the form 
The vertical axis scale is set by Vs = 4 cm /s
where → amplitude of oscillator
Vm = 5 cm/s since Vs= 4 cm/s
from the graph, at t = 0, v = 4 cm/s
If we take 
,acceleration would be positive but from the graph since the v is decreasing at
t = 0
∴ the acceleration should be negative!
∴ Ø = 53°
The correct answer is: 53
For a simple pendulum, find the angular displacement at which the restoring torque required for simple harmonic motion deviates from the actual restoring torque by 2% use g = 10m/s2 . (θ in degrees).
The restoring torque on the bob of the pendulum at an angle with the vertical is given by
We apply small angle approximation to so that the motion can be followed as simple harmonic motion, the torque becomes
The problem requires to find the value of 9 for which
Write the taylor expansion of Sin θ
Taking absolute value on both sides
The correct answer is: 19.65
A particle of mass m moves in the potential shown above find the period of the motion when the particle has energy E. (in terms of 
For the simple harmonic oscillator (SHO) part,
Since the graph shows only half of the usual SHO potential, the period contribution from the SHO part should be half the usual period
For the gravitational potential, one can calculate the period from the usual kinematics equation for constant acceleration
Since E is always conserved
At the end point 
x = Elm g
Since the particle has to travel from the origin to the right end point & then back again the total time contribution is twice
The correct answer is: 2
The circular hoops, X & Y are hanging on nails in a wall. The mass of X is four times of Yand the diameter of X is also four times that of Y and the diameter of X is also four times that of Y. If the period of small oscillations of Xis T, then find the period of oscillations for Y (in terms T)
Assuming both the loops to be point masses concentrated at the centre of mass, they can be considered as simple pendulums. For small oscillations, time period of a
simple pendulum 
Time period of Y is 
The correct answer is: 0.5
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