Answer to Question #174145 in Field Theory for Cj Santos

Question

You hang a ball using a 0.240m long string to a window. You pull it at an angle of 22.0° and then release it. a. how much time does it take for the ball to reach its highest speed? b. how much time does it take if the pendulum is released at an angle 11.0°

Answer 1

Answer

a) Time period is given

$T=2\pi\sqrt{\frac{l}{g}}\\=2\pi\sqrt{\frac{0.24}{9.8}}=0.98sec$

So angular velocity

$\omega=\frac{2\pi}{T}=6.39rad/sec$

So amplitude of oscillation

$A=l-lcos\theta=l-lcos\omega t=0.017$

$V=V_{max}sin\omega t\\=V_{max}=V_{max}sin\omega t$

Putting value of $\omega$

$6.39\times t=3.14/2$

t=0.25sec

b) When angel is 11° then then also time is equal to time in part (a) . Because that time is only depend on angular frequency and angular frequency is only depend on length of string. So time taken in t= 0.25s