Exercise 14.16 18 of 22 (> Part A Constants A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is 0.090 m, it takes the block 2.45 s to travel from 0.090 m to 0.180 m to0.180 m? f the amplitude is doubled, to 0.180 m, how long does it take the block to travel from Express your answer to two significant figures and include the appropriate units 0.090 m t-2.5 s Previous Answers Correct Part B f the amplitude is doubled, to 0.180 m, how long does it take the block to travel from Express your answer to two significant figures and include the appropriate units 0.090 m to -0.090 m Submit Previous Answers Request Answer X Incorrect; Try Again; 9 attempts remaining
Solution
Time required to travel from + maximum to - maximum is T/2, where T is the time period of oscillations T = 2 x 2.5 = 5
After doubling the amplitude, the distance from x = 0.09m to x = 0.09 m is time taken to travel from + half to - half of maximum amplitude T/6 = 0.833 seconds
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Exercise 14-16 18 of 22 (- Part A Constants A small block is attached.docx
1. Exercise 14.16 18 of 22 (> Part A Constants A small block is attached to an ideal spring and is
moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is 0.090
m, it takes the block 2.45 s to travel from 0.090 m to 0.180 m to0.180 m? f the amplitude is
doubled, to 0.180 m, how long does it take the block to travel from Express your answer to two
significant figures and include the appropriate units 0.090 m t-2.5 s Previous Answers Correct
Part B f the amplitude is doubled, to 0.180 m, how long does it take the block to travel from
Express your answer to two significant figures and include the appropriate units 0.090 m to -
0.090 m Submit Previous Answers Request Answer X Incorrect; Try Again; 9 attempts
remaining
Solution
Time required to travel from + maximum to - maximum is T/2, where T is the time period of
oscillations T = 2 x 2.5 = 5
After doubling the amplitude, the distance from x = 0.09m to x = 0.09 m is time taken to travel
from + half to - half of maximum amplitude T/6 = 0.833 seconds
