Question

Part II # 1 A mass on a string of unknown length oscillates as a pendulum with a period of 4 sec. What is the period if: (Parts (1-4 are independent questions, each referring to the initial situation.) 1) The mass is doubled? 2) The string length is doubled? 3) The string length is halved? 4) The amplitude is halved? A. 1) No Change (4 sec) 2) 2.8 sec 3) 5.7 sec 4) No Change (4 sec) B. 1) 5.7 sec 2) No Change (4 sec) 3) 2.8 sec 4) No Change (4 sec) C. 1) No Change (4 sec) 2) 5.7 sec 3) 2.8 sec 4) No Change (4 sec) D. 1) 5.7 sec 2) No Change (4 sec) 3) No Change (4 sec) 4) 2.8 sec

Answer 1

C. 1) No Change (4 sec) 2) 5.7 sec 3) 2.8 sec 4) No Change (4 sec)

Step-by-step explanation:

Given that:

Period (T) = 4 s

1) If the mass is doubled.

The period of a pendulum is given by the formula:

`T=2\pi\sqrt{(L)/(g) }` where L is the length and g is the acceleration due to gravity.

From the formula, the period does not depend on the mass of the spring therefore if the mass is doubled the period does not change.

2) The string length is doubled

Given that:

`T=2\pi\sqrt{(L)/(g) }, but\ T =4s\\4=2\pi\sqrt{(L)/(g) }`

if the length is doubled, the new spring length is 2L. Therefore the new period (T1) is given as:

`T_1=2\pi\sqrt{(2L)/(g) }=√(2) (2\pi\sqrt{(L)/(g) })=√(2)*4=5.7\ sec`

3) The string length is halved

Given that:

`T=2\pi\sqrt{(L)/(g) }, but\ T =4s\\4=2\pi\sqrt{(L)/(g) }`

if the length is halved, the new spring length is L/2. Therefore the new period (T1) is given as:

`T_1=2\pi\sqrt{(L)/(2g) }=√(1/2) (2\pi\sqrt{(L)/(g) })=√(1/2)*4=2.8\ sec`

4) The amplitude is halved

From the formula, the period does not depend on the amplitude therefore if the amplitude is halved the period does not change.