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)`.

, explain your 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

(this is obtained by solving the differential equation )

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

Since the angular frequency is , we can say that

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

= initial pressure

= final pressure = 2

= initial volume

= final volume = /4

= initial temperature = 20 c = 20 + 273 = 293 k

= final temperature = ?

using the equation