The essential formulae you have to remember in simple harmonic motion are the following:
(1) Equation of simple harmonic motion: y = Asinωt if initial phase and displacement are zero. Here ‘y’ is the displacement, ‘ω’ is the angular frequency and A is the amplitude.
y = Acosωt also represents simple harmonic motion but it has a phase lead of π/2 compared to the above one.
If there is an initial phase of Φ the equation is y = Asin(ωt + Φ).
y = Asinωt + Bcosωt represents the general simple harmonic motion of amplitude √(A2 + B2) and initial phase tan-1(B/A).
(2) The differential equation of simple harmonic motion is d2y/dt2 = -ω2y
Note that ω =√(k/m) where ‘k’ is the force constant (force per unit displacement) and ‘m’ is the mass of the particle executing the SHM.
(3) Velocity of the particle in SHM, v = ω√(A2 – y2)
Maximum velocity, vmax = ωA
(4) Acceleration of the particle in SHM, a = - ω2y
Maximum acceleration, amax = ω2A
(5) Kinetic Energy of the particle in SHM, K.E. = ½ m ω2( A2 –y2)
Maximum Kinetic energy = ½ m ω2A2
Potential Energy of the particle in SHM, P.E. = ½ m ω2y2
Maximum Potential Energy = ½ m ω2A2
Total Energy in any position = ½ m ω2A2
Note that the kinetic energy and potential energy are maximum respectively in the mean position and the extreme position. The sum of the kinetic and potential energies which is the total energy is a constant in all positions. Remember this:
Maximum K.E. = Maximum P.E. = Total Energy = ½ m ω2A2
(6) Period of SHM = 2π√(Inertia factor/ Spring factor)
In cases of linear motion as in the case of a spring-mass system or a simple pendulum, period, T = 2π √(m/k) where ‘m’ is the mass and ‘k’ is the force per unit displacement.
In the case of angular motion, as in the case of a torsion pendulum,
T = 2π √(I/c) where I is the moment of inertia and ‘c’ is the torque (couple) per unit angular displacement.
You may encounter questions requiring calculation of the period of seemingly difficult simple harmonic oscillators. Understand that the question will become simple once you are able to find out the force constant in linear motion and torque constant in angular motion. Angular cases will be rare in Medical and Engineering Entrance test papers. Let us now discuss some typical questions.
The following simple question appeared in the AIIMS 1998 test paper:
If a simple pendulum oscillates with an amplitude 50 mm and time period 2s, then its maximum velocity is
(a) 0.1 m/s (b) 0.15 m/s (c) 0.8 m/s (d) 0.16 m/s
Maximum velocity vmax = ωA where ‘ω’ is the angular frequency and ‘A’ is the amplitude. Therefore vmax = (2π/T)A = (2π/2)×50×10-3 = 0.157 m/s [Option (b)].
The following question appeared in Kerala Engineering Entrance 2005 test paper:
A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
(a) 1/ 2π√3 (b) 2π√3 (c) 2π/√3 (d) √3/2π (e) √3/π
The magnitudes of the velocity and acceleration of the particle when its displacement is ‘y’ are ω√(A2 –y2) and ω2y respectively. Equating them, ω√(A2 –y2) = ω2y, from which ω = [√(A2 –y2)]/y = √(4 –1) = √3. Period T = 2π/ω = 2π/√3.
Suppose you place a sphere of mass ‘m’ and radius ‘r’ inside a smooth, heavy hemispherical bowl of radius of 37r placed on a horizontal table. If the sphere is given a small displacement, what is its period of oscillation?
(a) 2π√(m/37rg) (b) 2π√(m/rg) (c) 12π√(r/g) (d) 2π√(r/g) (e) 2π√(37r/g)
The arrangement depicted in this question is similar to that of a simple pendulum. Instead of the usual string, you have a concave surface to confine the bob (sphere) to its path along the arc of a circle. The usual expression for the period, T = 2π√(L/g) holds here also, where the length L = 36r since the length of the pendulum is measured from the centre of gravity of the bob. The point of ‘suspension’ is evidently at the centre of the hemispherical bowl. The correct option is 12π√(r/g) given in (c).
What will be the period of oscillation of a simple pendulum of length 100 cm in a spaceship in a geostationary orbit?
(1) Equation of simple harmonic motion: y = Asinωt if initial phase and displacement are zero. Here ‘y’ is the displacement, ‘ω’ is the angular frequency and A is the amplitude.
y = Acosωt also represents simple harmonic motion but it has a phase lead of π/2 compared to the above one.
If there is an initial phase of Φ the equation is y = Asin(ωt + Φ).
y = Asinωt + Bcosωt represents the general simple harmonic motion of amplitude √(A2 + B2) and initial phase tan-1(B/A).
(2) The differential equation of simple harmonic motion is d2y/dt2 = -ω2y
Note that ω =√(k/m) where ‘k’ is the force constant (force per unit displacement) and ‘m’ is the mass of the particle executing the SHM.
(3) Velocity of the particle in SHM, v = ω√(A2 – y2)
Maximum velocity, vmax = ωA
(4) Acceleration of the particle in SHM, a = - ω2y
Maximum acceleration, amax = ω2A
(5) Kinetic Energy of the particle in SHM, K.E. = ½ m ω2( A2 –y2)
Maximum Kinetic energy = ½ m ω2A2
Potential Energy of the particle in SHM, P.E. = ½ m ω2y2
Maximum Potential Energy = ½ m ω2A2
Total Energy in any position = ½ m ω2A2
Note that the kinetic energy and potential energy are maximum respectively in the mean position and the extreme position. The sum of the kinetic and potential energies which is the total energy is a constant in all positions. Remember this:
Maximum K.E. = Maximum P.E. = Total Energy = ½ m ω2A2
(6) Period of SHM = 2π√(Inertia factor/ Spring factor)
In cases of linear motion as in the case of a spring-mass system or a simple pendulum, period, T = 2π √(m/k) where ‘m’ is the mass and ‘k’ is the force per unit displacement.
In the case of angular motion, as in the case of a torsion pendulum,
T = 2π √(I/c) where I is the moment of inertia and ‘c’ is the torque (couple) per unit angular displacement.
You may encounter questions requiring calculation of the period of seemingly difficult simple harmonic oscillators. Understand that the question will become simple once you are able to find out the force constant in linear motion and torque constant in angular motion. Angular cases will be rare in Medical and Engineering Entrance test papers. Let us now discuss some typical questions.
The following simple question appeared in the AIIMS 1998 test paper:
If a simple pendulum oscillates with an amplitude 50 mm and time period 2s, then its maximum velocity is
(a) 0.1 m/s (b) 0.15 m/s (c) 0.8 m/s (d) 0.16 m/s
Maximum velocity vmax = ωA where ‘ω’ is the angular frequency and ‘A’ is the amplitude. Therefore vmax = (2π/T)A = (2π/2)×50×10-3 = 0.157 m/s [Option (b)].
The following question appeared in Kerala Engineering Entrance 2005 test paper:
A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is
(a) 1/ 2π√3 (b) 2π√3 (c) 2π/√3 (d) √3/2π (e) √3/π
The magnitudes of the velocity and acceleration of the particle when its displacement is ‘y’ are ω√(A2 –y2) and ω2y respectively. Equating them, ω√(A2 –y2) = ω2y, from which ω = [√(A2 –y2)]/y = √(4 –1) = √3. Period T = 2π/ω = 2π/√3.
Suppose you place a sphere of mass ‘m’ and radius ‘r’ inside a smooth, heavy hemispherical bowl of radius of 37r placed on a horizontal table. If the sphere is given a small displacement, what is its period of oscillation?
(a) 2π√(m/37rg) (b) 2π√(m/rg) (c) 12π√(r/g) (d) 2π√(r/g) (e) 2π√(37r/g)
The arrangement depicted in this question is similar to that of a simple pendulum. Instead of the usual string, you have a concave surface to confine the bob (sphere) to its path along the arc of a circle. The usual expression for the period, T = 2π√(L/g) holds here also, where the length L = 36r since the length of the pendulum is measured from the centre of gravity of the bob. The point of ‘suspension’ is evidently at the centre of the hemispherical bowl. The correct option is 12π√(r/g) given in (c).
What will be the period of oscillation of a simple pendulum of length 100 cm in a spaceship in a geostationary orbit?
Well, in any satellite orbiting the earth (in any orbit), the condition of weightlessness exists (effective g = 0), the pendulum does not oscillate and the period therefore is infinite.
Consider the following question:
A simple pendulum is arranged using a small metallic bob of mass ‘m’and a light rubber cord of length ‘L’ (on suspending the bob), area of cross section ‘A’ and Young’s modulus ‘Y’. [One should use inextensible cord only for simple pendulum!]. When this unconventional pendulum is at rest in its mean position, the bob is pulled slightly down and is released. Then, the period of the vertical oscillation of the bob is (assuming that the size of the bob is negligible compared to the length of the cord)
(a) 2π√2L/g (b) 2π√(mL/YA) (c) 2π√ (m/YAL) (d) 2π√ (L/g) (e) 2π√ (mY/AL)
The period as usual is given by T = 2π√(m/k). Here ‘m’ is the same as the mass of the bob. The force constant can be found by writing the expression for Young’s modulus (since it arises from the elastic force in the cord): Y = FL/A(δL) where δL is the increase in the length of the cord on pulling the bob down with a force F. Therefore, the force constant, F/(δL) = YA/L. On substituting this value, the period is 2π√(mL/YA).
The following MCQ on simple harmonic motion may generate a little confusion in some of you:
A sphere of mass M is arranged on a smooth inclined plane of angle θ, in between two springs of spring constants K1 and K2 . The springs are joined to rigid supports on the inclined plane and to the sphere (Fig). When the sphere is displaced slightly, it executes simple harmonic motion. What is the period of this motion?
(a) 2π[Mgsinθ/(K1-K2)]½ (b) 2π[M/{K1K2/(K1+K2)}]½ (c) 2π[Mgsinθ/(K1+K2)]½ (d) 2π[M/(K1+K2)]½ (e) 2π[(K1+K2)/M]½
Consider the following question:
A simple pendulum is arranged using a small metallic bob of mass ‘m’and a light rubber cord of length ‘L’ (on suspending the bob), area of cross section ‘A’ and Young’s modulus ‘Y’. [One should use inextensible cord only for simple pendulum!]. When this unconventional pendulum is at rest in its mean position, the bob is pulled slightly down and is released. Then, the period of the vertical oscillation of the bob is (assuming that the size of the bob is negligible compared to the length of the cord)
(a) 2π√2L/g (b) 2π√(mL/YA) (c) 2π√ (m/YAL) (d) 2π√ (L/g) (e) 2π√ (mY/AL)
The period as usual is given by T = 2π√(m/k). Here ‘m’ is the same as the mass of the bob. The force constant can be found by writing the expression for Young’s modulus (since it arises from the elastic force in the cord): Y = FL/A(δL) where δL is the increase in the length of the cord on pulling the bob down with a force F. Therefore, the force constant, F/(δL) = YA/L. On substituting this value, the period is 2π√(mL/YA).
The following MCQ on simple harmonic motion may generate a little confusion in some of you:
A sphere of mass M is arranged on a smooth inclined plane of angle θ, in between two springs of spring constants K1 and K2 . The springs are joined to rigid supports on the inclined plane and to the sphere (Fig). When the sphere is displaced slightly, it executes simple harmonic motion. What is the period of this motion?
(a) 2π[Mgsinθ/(K1-K2)]½ (b) 2π[M/{K1K2/(K1+K2)}]½ (c) 2π[Mgsinθ/(K1+K2)]½ (d) 2π[M/(K1+K2)]½ (e) 2π[(K1+K2)/M]½
You should note that gravity has no effect on the period of oscillation of a spring-mass system since the restoring force is supplied by the elastic force in the spring. (It can oscillate with the same period in gravity free regions also). So, whether you place the system on an inclined plane or a horizontal plane, the period is the same and is determined by the effective spring constant and the attached mass only. The effective spring constant is K1 + K2 since both the springs try to enhance the opposition to the displacement of the mass. The period of oscillation, as usual is given by, T = 2π√(Inertia factor/Spring factor) = 2π√[M/(K1 + K2)], given in option (d).The following two questions (MCQ) appeared in Kerala Engineering Entrance 2006 test paper:
(1)The instantaneous displacement of a simple harmonic oscillator is given by y = A cos(ωt + π/4). Its speed will be maximum at the time
(a) 2π/ω (b) ω/2π (c) ω/π (d) π/4ω (e) π/ω
This question was omitted by a fairly bright student who got selected with a good rank. The question setter used the term speed (and not velocity) to make things very specific and to avoid the possible confusion regarding the sign. So what he meant is the maximum magnitude of velocity. The velocity is the time derivative of displacement: v = dy/dt = -Aω(sin ωt + π/4). Its maximum magnitude equal to Aω is obtained when ωt = π/4, from which t = π/4ω.
(2) A particle of mass 5 g is executing simple harmonic motion with an amplitude 0.3 m and time period π/5 s. The maximum value of the force acting on the particle is
(a) 5 N (b) 4 N (c) 0.5 N (d) 0.3 N (e) 0.15 N
If you remember the basic expression for period in the form, T = 2π√(m/k) where ‘k’ is the force constant, the solution becomes quite easy. From this, k = 4π2m/T2 = 4π2 ×5×10-3/(π/5)2 = 0.5. Since ‘k’ is the force for unit displacement, the maximum force is k times the maximum displacement (amplitude). Therefore maximum force = kA = 0.5×0.3 = 0.15N.
[If you remember that ω = √(k/m) you can arrive at the answer since T = 2π/ω].
(a) 5 N (b) 4 N (c) 0.5 N (d) 0.3 N (e) 0.15 N
If you remember the basic expression for period in the form, T = 2π√(m/k) where ‘k’ is the force constant, the solution becomes quite easy. From this, k = 4π2m/T2 = 4π2 ×5×10-3/(π/5)2 = 0.5. Since ‘k’ is the force for unit displacement, the maximum force is k times the maximum displacement (amplitude). Therefore maximum force = kA = 0.5×0.3 = 0.15N.
[If you remember that ω = √(k/m) you can arrive at the answer since T = 2π/ω].
Hello, the last question can be more easily done by using max a = w^2 * A
ReplyDeleteAlain
I thought of using the quantities already given in the question for finding the answer. Your suggestion is alright and can be found easier if you remember that the maximum acceleration, amax = ω^2×A. The maximum force is m×amax = mω^2×A where ω = 2π/T. Therefore, maximum force = m(2π/T)^2×A = 0.005×10^2×0.3 = 0.15 N.
ReplyDeleteGood morning Sir, Sir I need practice mcq topic wise can you give in pdf file.....???
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