As your jet plane speeds down the runway on takeoff, you measure its acceleration by suspending your yo-yo as a simple pendulum and noting that when the bob (of mass 725 g) is at rest relative to you, the string (of length 52.8 cm) makes an angle of 33° with the vertical. the acceleration of gravity is 9.81 m/s^2. find the period for small oscillations of this pendulum.

T = 1.46 s

Explanation:

The time period of pendulum can be given as:

------ (1)

Where,

L = Length of String = 52.8 cm

g = gravity = 9.81 m/s2

Substituting the values in equation (1),we get,

Notes:

  ------(2)

Where,

m = mass of object

k = Springs constant

The restoring force of pendulum which accelerates the mass towards equilibrium is given as:

mg Sinφ = ks

s = rφ

In this case, r = L

Therefore, s = Lφ

Now,

mg Sinφ = k Lφ

k = mg Sinφ / Lφ

k = (mg/L) (Sinφ/φ)

For all small angles:

lim (φ--->0) = Sinφ/φ = 1

therefore. equation (2) implies:

In this question: φ = 33°

33° = 33 x π /180

33° = 0.575959 radians

Therefore,

Sin 0.575959 / 0.575959

0.9456

1/0.9456 = 1.057505

Sqrt (1.057505)

1.02835

Now the formula can be written as:

So, the accurate answer would be 1.50 s.

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j