14.9k views
5 votes
If the spring constant is doubled, what value does the period have for a mass on a spring?

A. The period would double by square `sqrt(2)`.
B. The period would be halved by `sqrt(2)`.
C. The period would increase by `sqrt(2)`.
D. The period would decrease by `sqrt(2)`.

Please, explain your answer.

User JaKu
by
8.0k points

2 Answers

3 votes

Answer:

D. The period would decrease by `sqrt(2)`. is wrong on clever/plato

Step-by-step explanation:

i got it wrong :(

User Qqbt
by
6.9k points
7 votes

Answer:

The period would decrease by `sqrt(2)`.

Explanation:

The angular frequency omega of an oscillating mass m due to a spring with a constant k is given by


\omega = \sqrt{(k)/(m)}

(this is obtained by solving the differential equation
m\ddot{{ x}}-kx=0)

If k doubles, i.e., k'=2k, then


\omega'=\sqrt{(k')/(m)}=\sqrt{(2k)/(m)}=√(2)\sqrt{(k)/(m)}=\omega√(2)

Since the angular frequency is
\omega = (2\pi)/(T), we can say that


\omega√(2)=(2\pi)/(T)√(2)=(2\pi)/((T)/(√(2)))=(2\pi)/(T')

and so it becomes clear that the period T will decrease by sqrt(2) as stated in choice (D).

User Swivel
by
7.6k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.