P HYSICS F UNDAMENTAL F OR IIT-J EE
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
ω = L g ; f =
π ω 2 =
L g 2
1
π
T =
ω 2π
= f 1 = 2π
g L
A physical pendulum is a body suspended from an axis of rotation a distance d from its center of gravity.
If the moment of inertia about the axis of rotation is I, the angular frequency and period for small-amplitude oscillations are independent of amplitude.
ω = I
mgd ; T = 2π mgd
I
θ
mg sinθ
mg cosθ
mg
O z
d cg
d sinθ
Problem Solving Strategy : Simple Harmonic Motion I :
Step 1: Identify the relevant concepts : An oscillating system under goes simple harmonic motion (SHM) only if the restoring force is directly proportional to the displacement. Be certain that this is the case for the problem at hand before attempting to use any of the results of this section. As always, identify the target variables.
Step 2: Set up the problem using the following steps Identify the known and unknown quantities, and
determine which are the target variables.
It's useful to distinguish between two kinds of quantities. Basic properties of the system include the mass m and force constant k. (In some problems, m, k, or both can be determined from other information.) They also include quantities derived from m and k, such as the period T, frequency f, and angular frequency ω. Properties of the motion describe how the system behaves when it is set into motion in a particular way.
They include the amplitude A, maximum velocity vmax, and phase angle φ, as well as the values of x,vx, and ax at the particular time.
If necessary, define an x-axis as.
Step 3: Execute the solution as follows :
Use the equations T = 1/f and ω = 2πf = 2π/T to solve for the target variables.
If you need to calculate the phase angle, be certain to express it in radians. The quantity ωt in Eq. Fx = – kx is naturally in radians, so φ must be as well.
If you need to find the values of x, vx, and ax at various times, use Eqs.
f =
π ω 2 =
m k 2
1
π , vx = dt
dx = – ωA sin (ωt + φ)
and ax = dt dvx
= 2
2
dt x
d = –ω2A cos (ωt + φ).
If the initial position x0 and initial velocity v0x are both given, you can determine the phase angle and amplitude from Eqs. φ = arctan
ω0xx0
v
(phase angle in SHM) and A = x20 v202x + ω (amplitude in SHM). If the body is given an initial positive displacement x0 but zero initial velocity (v0x = 0), then the amplitude is A = x0
and the phase angle is φ = 0. If it has an initial positive velocity v0x but no initial displacement (x0 = 0), the amplitude is A = v0x / ω and the phase angle is φ = –π/2.
Step 4: Evaluate your answer : Check your results to make sure that they're consistent. As an example, suppose you've used the initial position and velocity to find general expressions for x and vx at time t. If you substitute t = 0 into these expressions, you should get back the correct values of x0 and vv x. Simple Harmonic Motion II
The energy equation E = mv2x
2
1 +
2 1kx2 =
2
1kA2 = constant ...(i) is a useful alternative relation between velocity and position, especially when energy quantities are also required. If the problem involves a relation among position, velocity, and acceleration without reference to time, it is usually easier to use Eq.
ax = 22 dt
x d = –
m
k x ...(ii)
(from Newton's second law) or eq. (i) (from energy conservation) than to use the general expressions for x, vx, and ax as functions of time [Eqs.
x = A cos (ωt + φ) (displacement in SHM), vx =
dt
dx = –ωA sin (ωt + φ) (velocity in SHM) and
ax = dt dvx
= 2
2
dt x
d = – ω2A cos (ωt + φ) (acceleration in SHM), respectively ]. Because the energy equation involves x2 and vx2, it cannot tell you the sign of x or vx, you have to infer the sign from the situation. For instance, if the body is moving from the equilibrium position towards the point of greatest positive displacement, then x is positive and vx is positive.
1. A body of mass 1 kg is executing simple harmonic motion which is given by x = 6.0 cos (100 t + π/4) cm. What is the (i) amplitude of displacement, (ii) frequency, (iii) initial phase, (iv) velocity, (v) acceleration, (vi) maximum kinetic energy ? Sol. The given equation of S.H.M. is
x = 6.0 cos (100 t + π/4) cm
Comparing it with the standard equation of S.H.M., x = a cos (ωt + φ), we have
(i) amplitude a = 6.0 cm (ii) frequency ω = 100 /sec (iii) initial phase φ = π/4
(iv) velocity v = ω (a2−x2) = 100 (36−x2) (v) acceleration = –ω2 x = – (100)2x = – 104 x (vi) kinetic energy =
2 1mv2 =
2
1mω2(a2 – x2) When x = 0, the kinetic energy is maximum, i.e., (K.E.)max =
2. A particle of mass 0.8 kg is executing simple harmonic motion with an amplitude of 1.0 metre and periodic time 11/7 sec. Calculate the velocity and the kinetic energy of the particle at the moment when its displacement is 0.6 meter.
Sol. We know that, v = ω (a2−y2)
Kinetic energy at this displacement is given by
K =
2 1mv2 =
2
1 × 0.8 × (3.2)2 = 4.1 joule 3. A person normally weighing 60 kg stands on a
platform which oscillates up and down harmonically at a frequency 2.0 sec–1 and an amplitude 5.0 cm. If a machine on the platform gives the person's weight against time, deduce the maximum and minimum reading it will show, take g = 10 m/sec2.
Sol. Acceleration of the platform a = ω2y Maximum acceleration
amax = ω2A (A = Amplitude)
4. A particle of mass m is located in a unidimensional potential field where the potential energy of the particle depends on the coordinate x as U(x) = U0(1 – cos C x); U0 and C are constants. Find the period of small oscillations that the particle performs about the equilibrium position.
Sol. Given that U(x) = U0(1 – cos C x) Here acceleration is directly proportional to the negative of displacement. So, the motion is S.H.M.
Time period T is given by
5. Find the period of small oscillations in a vertical plane performed by a ball of mass m = 40 g fixed at the middle of a horizontally stretched string l = 1.0 m in length. The tension of the string is assumed to be constant and equal to F = 10 N.
Sol. The situation is showing in fig. The components of T in upwards direction are T cos θ and T cos θ. Hence denominator.
∴ a = –
Here acceleration is directly proportional to the negative of displacement x. Hence the motion is S.H.M.
T = Substituting the given values, we get
T = 3.14 ×
Solved Examples
Preparation of Amines :
Through Nucleophilic Substitution Reactions - Alkylation of Ammonia Salts of primary amines can
be prepared from ammonia and alkyl halides by nucleophilic substitution reactions. Subsequent treatment of the resulting aminium salts with a base gives primary amines :
NH3 + R — X R — NH3 X– RNH2 OH–
•• +
This method is of very limited synthetic application because multiple alkylations occur.
A Mechanism for the Alkylation of NH3
NH3 + CH3CH2 — Br CH3CH2 — NH+ 3 + Br–
••
CH3CH2 — N — H + NH3 CH3CH2NH2 + NH4
•• +
H
H
••
CH3CH2NH•• 2+CH3CH2 —Br (CH3CH2)2NH+ 2+Br–, etc.
+
Alkylation of Azide ion and Reduction:
A much better method for preparing a primary amine from an alkyl halide is first to convert the alkyl halide to an alkyl azide (R—N3) by a nucleophilic substitution reaction:
R—X +•• –N=N=N+ –
••
•
•
•
• ( X ) 2 SN
−−
→
R–N=N=N•• + –•
•
•• LiAlHor 4
alcohol / Na
→
RN••H2
Azide ion Alkyl (A good nucleophile) azide
Then the alkyl azide can be reduced to a primary amine with sodium and alcohol or with lithium aluminum hydride. A word of caution: Alkyl azides are explosive, and low-molecular-weight alkyl azides should not be isolated but should be kept in solution.
Sodium azide is used in automotive airbags.
The Gabriel Synthesis:
Potassium phthalimide (see the following reaction) can also be used to prepare primary amines by a method known as the Gabriel synthesis. This synthesis also avoids the complications of multiple alkylations that occur when alkyl halides are treated with ammonia:
N—H O