HW16 ( SHM and Wave motion) Begin Date: 11/16/2017 12:01:00 AM Due Date: 11/28/2
ID: 1794646 • Letter: H
Question
HW16 ( SHM and Wave motion) Begin Date: 11/16/2017 12:01:00 AM Due Date: 11/28/2017 11:59:00 PM End Date: 12/14/201711:5900 PM (796) Problem 9: A mass m is oscillating with amplitude A at the end of a vertical spring of spring constant k 17% Part (a) The mass is increased by a factor of four, what is true about the period? Choose the best answer. The period decreases by a factor of four O The period increases by a factor of four. O The period doubles. OThe period is halved. Grade Summary Potential 100% Submissions Attempts remaining 3 0% 3% per attempt) tailed view Hint 1 give upt Hints: deduction per hint Hints remaining Feedback: deduction per feedback. 1796 Part (b) The mass is increased by a factor of four, keeping A and k constant what is true about the total mechanical energy? Choose 17% Part (c) The mass is increased by a factor of four, keeping A and k constant what is true about the maximum speed? Choose the best 17% Part (d) The amplitude is doubled, keeping mass and k constant. what is true about the penod? Choose the best answer. the best answer answer 1796 Part (e) The amplitude is doubled, keeping mass and k constant. what is true about the total mechanical energy? Choose the best answer ii 17% Part (f) The anplitude is doubled, keeping mass and k constant what is true about the maximum speed? Choose the best answer.Explanation / Answer
(A) T = 2pi sqrt(m/k)
m -> 4m
then T -> 2 T
Ans : Period doubles
(B) ME = k A^2 / 2
k and A constant.
Mechanical energy will remain same.
(C) k A^2 /2 = m v_max^2 /2
v_max = A sqrt(k / m)
m -> 4m
v_max -> 1/2 v_max
halved.
(d) period is independent of A .
will remain same.
(e) ME = k A^2 / 2
A - > 2 A
them ME = 4 ME
increases by a fctor of four.
(f) v_max also doubled.
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