Question

A pendulum on Planet X, where the value of m is unknown, oscillates with a period T1 = 8 s. Note that you do not know the value of m, L, or g, so do not assume any specific values. The required analysis involves thinking about ratios.
(A) What is the period of this pendulum if its mass is doubled? Express your answer with the appropriate units.
(B) Its length is doubled? Express your answer with the appropriate units.
(C) Its oscillation amplitude is doubled? Express your answer with the appropriate units.

Answer 1

Answer:

Explanation:

Given:

T1 = 8 s

For a simple harmonic motion, period, T = 2pi × sqrt(L/g)

A.

M2 = 2 × M1

The period is not dependent on mass.

T = 2pi × sqrt(L/g)

T1 = T2 = 8 s

B.

L2 = 2 × L1

T = 2pi × sqrt(L/g)

L1/T1^2 = L2/T2^2

T2 = sqrt((2 × L1 × 64)/L1)

= sqrt(128)

= 11.314 s

C.

Period is independent on the amplitude of the oscillation. T1 = T2 = 8 s

Answer 2

(a) When the mass of the pendulum is doubled, the period is 8 s.

(b) When the length of the pendulum is doubled, the period of oscillation is 11.3 s.

(c) When the amplitude of oscillation is doubled, the period is 8 s.

The period of oscillation of a pendulum is calculated as follows;

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

The period of the oscillation is independent of the mass, when the mass is doubled, the period is 8 s.

When the length is doubled, the period becomes;

$T_2 =2\pi \sqrt{\frac{2l}{g} } \\\\T_2 \ = \sqrt{2}\times T_1\\\\T_2 = 11.3 \ s$

The period of the oscillation is independent of the amplitude of the oscillation, when the amplitude is doubled, the period is 8 s.

Learn more about period of oscillation here: https://brainly.com/question/14596666