Class 11 Physics Notes Chapter 12 (Chapter 12) – Lab Manual (English) Book
Right then, let's get straight into the key concepts from Chapter 12 of your Physics Lab Manual, focusing on what's important for your government exam preparation. This chapter primarily deals with experiments related to Simple Harmonic Motion (SHM), specifically focusing on the Simple Pendulum and the Spring-Mass System.
Chapter 12: Simple Harmonic Motion (Simple Pendulum & Spring-Mass System)
Core Concept: Simple Harmonic Motion (SHM)
- Definition: SHM is a specific type of periodic motion (motion that repeats itself at regular intervals) where the restoring force acting on the object is directly proportional to its displacement from the mean (equilibrium) position and is always directed towards that mean position.
- Mathematical Representation:
- Restoring Force:
F = -kx
F
= Restoring Forcek
= Force constant (a positive constant specific to the system, like spring constant or equivalent)x
= Displacement from the mean position- The negative sign indicates the force opposes the displacement.
- Acceleration:
a = F/m = -(k/m)x
- Since
k/m
is constant for a given system, we writea = -ω²x
, whereω = √(k/m)
is the angular frequency.
- Since
- Restoring Force:
- Key Condition: The defining characteristic of SHM is that the acceleration is directly proportional to the negative of the displacement (
a ∝ -x
).
1. The Simple Pendulum
- Setup: An idealized model consisting of a point mass (called the 'bob') suspended by a massless, inextensible string from a rigid support. In practice, we use a small, heavy bob and a light string.
- Motion: When displaced slightly from its vertical equilibrium position and released, it oscillates. For small angular displacements (typically less than 10-15 degrees), its motion is approximately SHM.
- Why small amplitude? The restoring force is
F = -mg sinθ
. For small angles (measured in radians),sinθ ≈ θ
. Also, the arc lengthx = Lθ
, soθ = x/L
. Substituting this givesF ≈ -mg(x/L) = -(mg/L)x
. This fits theF = -kx
form, with the effective force constantk = mg/L
. If the angle is large,sinθ ≠ θ
, and the motion is periodic but not simple harmonic. - Time Period (T): The time taken for one complete oscillation (e.g., from one extreme point back to the same extreme point).
- Formula:
T = 2π√(L/g)
L
= Effective Length: This is crucial. It's the distance from the point of suspension to the centre of gravity of the bob (Length of string + radius of the bob).g
= Acceleration due to gravity at that location.
- Formula:
- Frequency (f): Number of oscillations per second.
f = 1/T = (1/2π)√(g/L)
- Factors Affecting Time Period (T):
- Directly Proportional to the square root of the effective length (
T ∝ √L
). Longer pendulum, longer period. - Inversely Proportional to the square root of acceleration due to gravity (
T ∝ 1/√g
). Pendulum oscillates slower on the Moon (where g is less) than on Earth. - Independent of the mass of the bob. A heavy bob and a light bob of the same length will have the same time period.
- Independent of the amplitude of oscillation, provided the amplitude is small.
- Directly Proportional to the square root of the effective length (
- Experimental Determination of 'g':
- Measure time (
t
) forN
(e.g., 20 or 30) oscillations for various effective lengths (L
). CalculateT = t/N
. - Plot a graph of
T²
(y-axis) vsL
(x-axis). - According to the formula
T = 2π√(L/g)
, squaring both sides givesT² = (4π²/g)L
. - This is in the form
y = mx
, so the graph should be a straight line passing through the origin. - The slope of the
T²
vsL
graph ism = 4π²/g
. - Therefore,
g = 4π² / slope
. This is how 'g' can be determined experimentally.
- Measure time (
- Key Graphs:
T
vsL
: Parabolic curve opening towards the L-axis.T²
vsL
: Straight line passing through the origin.
- Important Precautions/Sources of Error:
- Support must be rigid.
- Amplitude must be kept small.
- Measure effective length carefully to the C.G. of the bob.
- Count a large number of oscillations (
N
) to minimize error in measuringT
. - Ensure the pendulum oscillates in a single vertical plane (no spinning or elliptical motion).
- Minimize air resistance (use a dense, spherical bob).
2. Spring-Mass System
- Setup: A mass (
m
) attached to a light helical spring, which can oscillate vertically or horizontally. - Hooke's Law: Within the elastic limit, the force (
F
) exerted by the spring (restoring force) is proportional to the extension or compression (x
) from its equilibrium position:F = -kx
.k
= Spring Constant (or force constant). Unit: N/m. It measures the stiffness of the spring. A higherk
means a stiffer spring.
- Motion: The oscillations of the mass are SHM.
- Time Period (T):
- Formula:
T = 2π√(m/k)
m
= Mass attached to the spring. (For high accuracy, the effective massm_eff = m + m_spring/3
should be used, but often the spring mass is neglected).k
= Spring constant.
- Formula:
- Frequency (f):
f = 1/T = (1/2π)√(k/m)
- Factors Affecting Time Period (T):
- Directly Proportional to the square root of the mass (
T ∝ √m
). More mass, longer period. - Inversely Proportional to the square root of the spring constant (
T ∝ 1/√k
). Stiffer spring (larger k), shorter period. - Independent of acceleration due to gravity (
g
) for horizontal oscillations. For vertical oscillations,g
determines the equilibrium position (mg = kx₀
), but the period of oscillation around this position still follows the formulaT = 2π√(m/k)
.
- Directly Proportional to the square root of the mass (
- Experimental Determination of 'k':
- Method 1 (Static - Hooke's Law): Apply different known forces (e.g., by hanging weights,
F = Mg
) and measure the corresponding extensions (x
) from the unloaded position. PlotF
(y-axis) vsx
(x-axis). The graph should be a straight line passing through the origin. The slope of this graph is the spring constantk
. - Method 2 (Dynamic - Oscillations): Attach different known masses (
m
) to the spring, displace them slightly, and measure the time period (T
) of vertical oscillations (time forN
oscillations /N
). PlotT²
(y-axis) vsm
(x-axis). - From
T = 2π√(m/k)
, we getT² = (4π²/k)m
. - This is the form
y = (slope) * x
. The graph should be a straight line. - The slope of the
T²
vsm
graph isslope = 4π²/k
. - Therefore,
k = 4π² / slope
. The value ofk
obtained dynamically should ideally match the value obtained statically.
- Method 1 (Static - Hooke's Law): Apply different known forces (e.g., by hanging weights,
- Key Graphs:
F
vsx
(Load vs Extension): Straight line through the origin (slope = k).T
vsm
: Parabolic curve opening towards the m-axis.T²
vsm
: Straight line (ideally through origin, but may have a small positive intercept on T² axis if spring mass is significant).
- Important Precautions/Sources of Error:
- Ensure loading is within the elastic limit of the spring.
- Measure extensions carefully, avoiding parallax error.
- Ensure purely vertical oscillations without swinging.
- Use a stopwatch accurately and count a large number of oscillations.
- Use a rigid support.
Multiple Choice Questions (MCQs)
Here are 10 MCQs based on the concepts discussed, suitable for exam practice:
-
The time period of a simple pendulum depends on:
(a) Mass of the bob
(b) Amplitude of oscillation
(c) Length of the pendulum
(d) Size of the bob -
If the length of a simple pendulum is increased by a factor of 4, its time period will:
(a) Increase by a factor of 2
(b) Decrease by a factor of 2
(c) Increase by a factor of 4
(d) Remain unchanged -
A simple pendulum is taken from the equator to the poles. Its time period will:
(a) Increase
(b) Decrease
(c) Remain the same
(d) Become infinite
(Hint: How does 'g' change from equator to poles?) -
In the experiment to determine 'g' using a simple pendulum, the graph of T² versus L is a:
(a) Parabola
(b) Hyperbola
(c) Straight line passing through the origin
(d) Straight line with a non-zero intercept -
A mass 'm' is suspended from a spring of force constant 'k'. The time period of vertical oscillation is T. If the mass is doubled to 2m, the new time period will be:
(a) T/2
(b) T/√2
(c) √2 T
(d) 2T -
The force constant 'k' of a spring can be determined from the slope of the graph between:
(a) Load (F) and (Extension)²
(b) Load (F) and Extension (x)
(c) (Time Period)² and Mass (m)
(d) Time Period (T) and Mass (m) -
For a spring-mass system oscillating vertically, the time period depends on:
(a) Acceleration due to gravity (g)
(b) Amplitude of oscillation
(c) Mass (m) and Spring constant (k)
(d) Both (a) and (c) -
Two springs have force constants k₁ and k₂ such that k₁ > k₂. If they are stretched by the same force, then:
(a) Extension in spring 1 (x₁) > Extension in spring 2 (x₂)
(b) Extension in spring 1 (x₁) < Extension in spring 2 (x₂)
(c) x₁ = x₂
(d) Cannot be determined -
The condition
a = -ω²x
is characteristic of:
(a) Any periodic motion
(b) Uniform circular motion
(c) Simple Harmonic Motion
(d) Damped oscillation -
To minimize error in measuring the time period of a simple pendulum, one should:
(a) Use a very short string
(b) Use a very large amplitude
(c) Measure the time for only one oscillation
(d) Measure the time for a large number (e.g., 20 or 30) of oscillations
Answer Key for MCQs:
- (c)
- (a) (Since T ∝ √L, if L becomes 4L, T becomes √4 T = 2T)
- (b) ('g' is maximum at the poles and minimum at the equator. Since T ∝ 1/√g, T decreases at the poles)
- (c)
- (c) (Since T ∝ √m, if m becomes 2m, T becomes √2 T)
- (b) (Slope of F vs x graph is k)
- (c)
- (b) (Since F = kx, x = F/k. For the same F, larger k means smaller x)
- (c)
- (d)
Study these points thoroughly, focusing on the formulae, the dependencies of the time period, and the interpretation of graphs. Understanding the sources of error and precautions is also beneficial for conceptual clarity. Good luck with your preparation!