The starting angle θθ of a pendulum does not affect its period for θ<<1θ<<1. At higher angles, however, the period TT increases with increasing θθ.
The relation between TT and θθ can be derived by solving the equation of motion of the simple pendulum (from F=ma)
−gsinθ=lθ¨−gainθ=lθ¨
For small angles, θ≪1,θ≪1, and hence sinθ≈θsinθ≈θ. Hence,
θ¨=−glθθ¨=−glθ
This second-order differential equation can be solved to get θ=θ0cos(ωt),ω=gl−−√θ=θ0cos(ωt),ω=gl. The period is thus T=2πω=2πlg−−√T=2πω=2πlg, which is independent of the starting angle θ0θ0.
For large angles, however, the above derivation is invalid. Without going into the derivation, the general expression of the period is T=2πlg−−√(1+θ2016+...)T=2πlg(1+θ0216+...). At large angles, the θ2016θ0216 term starts to grow big and cause
Answer:
Explanation:
Given
Height from which cart is released is ![h=10\ m](https://tex.z-dn.net/?f=h%3D10%5C%20m)
as we know energy is Conserved and changes only its form so Potential Energy of Cart converted to kinetic Energy at bottom
Energy at Top ![E_T=mgh](https://tex.z-dn.net/?f=E_T%3Dmgh)
where m=mass of cart
g=acceleration due to gravity
h=height of track w.r.t bottom
Energy at bottom ![E_B=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=E_B%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
![E_T=E_B](https://tex.z-dn.net/?f=E_T%3DE_B)
![mgh=\frac{1}{2}mv^2](https://tex.z-dn.net/?f=mgh%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
cancel out common terms
![v=\sqrt{2\times g\times h}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B2%5Ctimes%20g%5Ctimes%20h%7D)
![v=\sqrt{2\times 9.8\times 10}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B2%5Ctimes%209.8%5Ctimes%2010%7D)
![v=14\ m/s](https://tex.z-dn.net/?f=v%3D14%5C%20m%2Fs)
There would be more hurricanes because hurricanes from by water.
More rain because you would get more precipitation.
Higher flooding if a storm comes through.
Complete Question
One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a small patch of light on the far wall. Having recently studied optics in your physics class, you're not too surprised to see that the patch of light seems to be a circular diffraction pattern. It appears that the central maximum is about 2 cm across, and you estimate that the distance from the window shade to the wall is about 5 m.
Required:
Estimate the diameter of the pinhole.
Answer:
The diameter is ![d =0.000336 m](https://tex.z-dn.net/?f=d%20%3D0.000336%20m)
Explanation:
From the question we are told that
The central maxima is ![D= 2cm = \frac{2}{100} = 0.02m](https://tex.z-dn.net/?f=D%3D%202cm%20%3D%20%5Cfrac%7B2%7D%7B100%7D%20%3D%200.02m)
The distance from the window shade is ![L = 5m](https://tex.z-dn.net/?f=L%20%3D%205m)
The average wavelength of the sun is mathematically evaluated as
![\lambda_{ave } = \frac{\lambda_i + \lambda_f}{2}](https://tex.z-dn.net/?f=%5Clambda_%7Bave%20%7D%20%3D%20%5Cfrac%7B%5Clambda_i%20%20%2B%20%5Clambda_f%7D%7B2%7D)
Generally the visible light spectrum has a wavelength range between 400 nm to 700 nm
So the initial wavelength of the sun is ![\lambda _i = 400nm](https://tex.z-dn.net/?f=%5Clambda%20_i%20%3D%20400nm)
and the final wavelength is ![\lambda_f = 700nm](https://tex.z-dn.net/?f=%5Clambda_f%20%3D%20700nm)
Substituting this into the above equation
![\lambda_{sun} = \frac{400nm +700nm}{2}](https://tex.z-dn.net/?f=%5Clambda_%7Bsun%7D%20%3D%20%5Cfrac%7B400nm%20%20%2B700nm%7D%7B2%7D)
![= 550nm](https://tex.z-dn.net/?f=%3D%20550nm)
The diameter is evaluated as
![d = \frac{2.44 \lambda_{sun} L}{D}](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B2.44%20%5Clambda_%7Bsun%7D%20L%7D%7BD%7D)
substituting values
![d = \frac{2.44 * 550*10^{-9} * 5 }{0.02}](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7B2.44%20%2A%20550%2A10%5E%7B-9%7D%20%2A%205%20%7D%7B0.02%7D)
![d =0.000336 m](https://tex.z-dn.net/?f=d%20%3D0.000336%20m)