Question

A pendulum of length 130.0 cm has a period of oscillation, T₁. The bob is pulled and released to move in a horizontal circle of radius 50.0 cm. If the period of rotation is T2, calculate the ratio T₁: T2. [g=10m s?, n=3.142]

Answer 1

The ratio of T1 : T2 is 1.6

What is Simple Harmonic Motion ?

SHM is the periodic motion of a particle along a line such that the acceleration of the body is directed toward a fixed point and proportional to its displacement.

Given that a pendulum of length 130.0 cm has a period of oscillation, T₁.

Then, The period T₁ = 2$\pi$$\sqrt{\frac{l}{g} }$

T₁ = 2 x 3.142 $\sqrt{\frac{1.3}{10} }$

T₁ = 6.284 x 0.36

T₁ = 2.266

But if T = 0

then, m$v^{2}$ /r = mg

the m will cancel out

Then v = $\sqrt{rg}$

where r = 50/100 = 0.5m

Substitute into the formula

v = $\sqrt{ 0.5 * 10}$

v = 2.24 m/s

Also, given that the bob is pulled and released to move in a horizontal circle of radius 50.0 cm and the period of rotation is T2,

w = 2$\pi$f

w = 2$\pi$/T

but v = wr

w = v/r

Therefore,

v/r = 2$\pi$/T

Substitute all the parameters into the formula

2.24/0.5 = (2 x 3.1423) /T

T = 6.284/4.472

T = 1.405 = $T_{2}$

The ratio T₁: T2 will be 2.266/1.405

The ratio is 1.6

Learn more about Simple Harmonic Motion here: https://brainly.com/question/24646514

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