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Trigonometric functions around a unit circle
26/02/2026

Trigonometric functions around a unit circle

Omar Yaghi, a chemist at the University of California, Berkeley and 2025 Nobel Prize winner, has developed a machine tha...
25/02/2026

Omar Yaghi, a chemist at the University of California, Berkeley and 2025 Nobel Prize winner, has developed a machine that can produce up to 1,000 liters of clean drinking water per day from air.

Built by his company Atoco, the device works even in very dry areas with less than 20% humidity. It runs on low-grade heat, such as sunlight, so it can operate off-grid without relying on electricity.

The system uses advanced materials called Metal-Organic Frameworks (MOFs). These porous materials act like powerful sponges, capturing water molecules from the air. When gently heated, they release the moisture, which is then condensed into clean liquid water.

About the size of a 20-foot shipping container, the unit can be placed in remote desert villages, islands, or disaster-hit regions. It could provide reliable drinking water in areas affected by droughts or hurricanes, especially when central water and power systems are damaged.

The technology is based on Yaghi’s Nobel Prize-winning work in reticular chemistry, which made MOFs possible and opened the door to applications like water harvesting and carbon capture.

16/02/2026

➡️ Period of a Simple Pendulum & Laws
The time taken by a simple pendulum to complete one full oscillation (i.e., to swing from its initial position to the opposite extreme and back to the same position).

➡️ Derivation
Consider:
Length of pendulum = L
Mass of bob = m
Angular displacement = θ (in radians)
Acceleration due to gravity = g
When displaced by a small angle θ, a restoring force acts toward the mean position.

Restoring force along arc = -mg sinθ
(The negative sign indicates the force acts toward equilibrium.)

For small angles (θ less than about 15°):
sinθ ≈ θ (in radians)

So,
Restoring force ≈ -mgθ
Since arc length s = Lθ,
θ = s / L

Substitute:
Restoring force = -mg (s/L)
= -(mg/L) s
Using Newton’s Second Law:
F = m d²s/dt²
So,
m d²s/dt² = -(mg/L) s

Divide by m:
d²s/dt² + (g/L) s = 0
This is the standard equation of simple harmonic motion (SHM):
d²x/dt² + ω²x = 0

Comparing:
ω² = g/L

So,
ω = √(g/L)
We know,
T = 2π / ω

Therefore,
T = 2π √(L/g)
This is the fundamental formula of a simple pendulum.

Laws os simple pendulum
➡️ Law of Isochronism
Definition:
For small oscillations, the time period is independent of amplitude.
From the formula:
T = 2π √(L/g)

There is no amplitude term in this equation.
Hence, for small angles:
Time period does not depend on amplitude.
Note: For large angles, this law does not hold exactly.

➡️ Law of Length
Statement:
Time period is directly proportional to the square root of the length.

From:
T = 2π √(L/g)
Since 2π and g are constants:
T ∝ √L
If length becomes 4 times:
New T = 2π √(4L/g)
= 2 × 2π √(L/g)
So time period doubles.

➡️ Law of Mass
From:
T = 2π √(L/g)
Mass (m) does not appear in the formula.
Therefore:
Time period is independent of mass of the bob.
A heavy and light bob of same length oscillate in equal time.

➡️ Law of Gravity
From:
T = 2π √(L/g)
T ∝ 1/√g
If gravity decreases:
Time period increases.

Example:
On Moon, g ≈ 1.62 m/s²
On Earth, g ≈ 9.8 m/s²

Since Moon has smaller g:
Time period on Moon > Time period on Earth.

➡️ Frequency and Angular Frequency
Frequency (f):
f = 1/T
= 1 / [2π √(L/g)]
= (1/2π) √(g/L)

Angular frequency (ω):
ω = √(g/L)

Energy of a Pendulum:
Total mechanical energy remains constant.
At extreme position:
Kinetic Energy (KE) = 0
Potential Energy (PE) = maximum

At mean position:
KE = maximum
PE = minimum

Maximum potential energy at angle θ:
PE = mgL (1 − cosθ)

For small θ:
cosθ ≈ 1 − θ²/2
So,
PE ≈ (1/2) mgL θ²
This shows pendulum motion behaves like SHM for small angles.

➡️ Period of a Simple Pendulum & Laws The time taken by a simple pendulum to complete one full oscillation (i.e., to swi...
16/02/2026

➡️ Period of a Simple Pendulum & Laws
The time taken by a simple pendulum to complete one full oscillation (i.e., to swing from its initial position to the opposite extreme and back to the same position).

➡️ Derivation
Consider:
Length of pendulum = L
Mass of bob = m
Angular displacement = θ (in radians)
Acceleration due to gravity = g
When displaced by a small angle θ, a restoring force acts toward the mean position.

Restoring force along arc = -mg sinθ
(The negative sign indicates the force acts toward equilibrium.)

For small angles (θ less than about 15°):
sinθ ≈ θ (in radians)

So,
Restoring force ≈ -mgθ
Since arc length s = Lθ,
θ = s / L

Substitute:
Restoring force = -mg (s/L)
= -(mg/L) s
Using Newton’s Second Law:
F = m d²s/dt²
So,
m d²s/dt² = -(mg/L) s

Divide by m:
d²s/dt² + (g/L) s = 0
This is the standard equation of simple harmonic motion (SHM):
d²x/dt² + ω²x = 0

Comparing:
ω² = g/L

So,
ω = √(g/L)
We know,
T = 2π / ω

Therefore,
T = 2π √(L/g)
This is the fundamental formula of a simple pendulum.

Laws os simple pendulum
➡️ Law of Isochronism
Definition:
For small oscillations, the time period is independent of amplitude.
From the formula:
T = 2π √(L/g)

There is no amplitude term in this equation.
Hence, for small angles:
Time period does not depend on amplitude.
Note: For large angles, this law does not hold exactly.

➡️ Law of Length
Statement:
Time period is directly proportional to the square root of the length.

From:
T = 2π √(L/g)
Since 2π and g are constants:
T ∝ √L
If length becomes 4 times:
New T = 2π √(4L/g)
= 2 × 2π √(L/g)
So time period doubles.

➡️ Law of Mass
From:
T = 2π √(L/g)
Mass (m) does not appear in the formula.
Therefore:
Time period is independent of mass of the bob.
A heavy and light bob of same length oscillate in equal time.

➡️ Law of Gravity
From:
T = 2π √(L/g)
T ∝ 1/√g
If gravity decreases:
Time period increases.

Example:
On Moon, g ≈ 1.62 m/s²
On Earth, g ≈ 9.8 m/s²

Since Moon has smaller g:
Time period on Moon > Time period on Earth.

➡️ Frequency and Angular Frequency
Frequency (f):
f = 1/T
= 1 / [2π √(L/g)]
= (1/2π) √(g/L)

Angular frequency (ω):
ω = √(g/L)

Energy of a Pendulum:
Total mechanical energy remains constant.
At extreme position:
Kinetic Energy (KE) = 0
Potential Energy (PE) = maximum

At mean position:
KE = maximum
PE = minimum

Maximum potential energy at angle θ:
PE = mgL (1 − cosθ)

For small θ:
cosθ ≈ 1 − θ²/2
So,
PE ≈ (1/2) mgL θ²
This shows pendulum motion behaves like SHM for small angles.

Heat transfer
09/02/2026

Heat transfer

Kinetic Theory of Gases, BSc HRK Physics Notes
05/02/2026

Kinetic Theory of Gases, BSc HRK Physics Notes

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