Question

Question: The period of a simple pendulum is T=2π√(L/g) with L = length of the pendulum and g = gravity of earth.

a.Derive the scaling law for the pendulum.

b.How much would the period of the pendulum change when you decrease the characteristic dimension by a factor of 10?

Answer 1

A) the scaling law is the relationship between two physical quantities. As here we have given the relationship between the time and length as,

Here factor 2π contain the all other dimensional variable as radius of Bob r, mass of length, l. As other factor define in F then,

B) here if L decrease by factor of 10 the L= 0.1L then T will going to decrease by the factor of 3.16.

Answer 2

Let's break down this pendulum problem:

a. Derive the Scaling Law for the Pendulum

The period (T) of a simple pendulum is given by:

  T = 2π√(L/g)

where:

• T is the period (time for one complete oscillation)

• L is the length of the pendulum

• g is the acceleration due to gravity (assumed constant)

We want to find how the period (T) scales with the length (L).

Let's rewrite the equation to isolate the length dependence:

T = 2π * (L/g)^(1/2)

T = 2π * (1/√g) * L^(1/2)

Since 2π and g are constants, we can write:

T ∝ √L

This is the scaling law for the pendulum: The period (T) is proportional to the square root of the length (L).

b. How much would the period of the pendulum change when you decrease the characteristic dimension (length) by a factor of 10?

Let's say the initial length is L₁. Then the initial period T₁ is:

T₁ = 2π√(L₁/g)

Now, we decrease the length by a factor of 10, so the new length L₂ is:

L₂ = L₁ / 10

The new period T₂ is:

T₂ = 2π√(L₂/g)

T₂ = 2π√((L₁/10) / g)

T₂ = 2π√(L₁ / (10g))

T₂ = 2π * √(L₁/g) / √10

T₂ = T₁ / √10

Therefore, the new period T₂ is the initial period T₁ divided by √10.

Calculate the change:

√10 ≈ 3.16

So, T₂ ≈ T₁ / 3.16

Answer:

The period of the pendulum would decrease by a factor of √10 (approximately 3.16) when the length is decreased by a factor of 10.

Answer 3

a). Derive the Scaling Law for the Pendulum :-

The period $T$ of a simple pendulum is given by:

$$T=2\pi \sqrt{\frac{L}{g}}$$

where:

• $L$ = length of the pendulum,

• $g$ = acceleration due to gravity (constant on Earth).

Scaling Law:
The period $T$ scales with the square root of the length $L$:

$$T\propto \sqrt{L}$$

This means:

• If the length $L$ increases, the period $T$ increases, but only by the square root of the change in length.

• Similarly, if $L$ decreases, $T$ decreases by the square root of the change.

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b). Change in Period When Length is Reduced by a Factor of 10 :-

If the original length is $L$, the new length becomes $\frac{L}{10}$.

Using the scaling law:

$$T_{\text{new}}=2\pi \sqrt{\frac{L\mathrm{/}10}{g}}=\frac{1}{\sqrt{10}}\cdot 2\pi \sqrt{\frac{L}{g}}=\frac{T_{\text{original}}}{\sqrt{10}}$$$$\sqrt{10}\approx 3.162$$

So:

$$T_{\text{new}}\approx \frac{T_{\text{original}}}{3.162}$$

Conclusion:

• The period decreases by a factor of $\sqrt{10}$ (≈ 3.162) when the length is reduced by a factor of 10.

• In other words, the pendulum swings faster (shorter period) when made shorter.

For example:

• If the original period was $2$ seconds, the new period would be $\frac{2}{3.162}\approx 0.632$$\frac{2}{3.162}\approx$$0.632$ seconds.