A block attached to a spring with unknown spring constant oscillates with a period of 2.0 s. What is the period if c. The amplitude is doubled?

Video transcript

Hey, everyone in this problem, an ideal spring with a mass of two kg attached to one of its ends rests on a smooth horizontal surface. The other end is fixed to a support. The spring constant is K when the spring is stretched by five centimeters, the mass oscillates with a period of three seconds. We're asked to calculate the oscillation period. If the spring is stretched by 7.5 centimeters, we're given four answer choices. Option A two seconds, option B three seconds, option C 4.5 seconds and option D 11.3 seconds. Now, let's go ahead and write out the information we're given. So we're told that the mass is equal to two kg. We're told that when the string is spr stretched by or sorry, the spring is stretched by five centimeters. The mass oscillates with a period of three seconds. So we're gonna call this 0.1. OK. So X one is gonna be this five centimeter stretch and we know that the period when we have this stretch, T one is equal to three seconds. And what we're asked to do is to calculate T two when the stretch is 7.5 centimeters. OK. So when X two is equal to 7.5 cents, now recall that the period T is given by two pi multiplied by the square root of M divided by K. Now, when we look at this equation for period, we can see that it only depends on the mass M and the spring constant K. OK. Neither of those change by stretching the spring. OK. The spring constant is inherent. It's a constant. It's the same for that spring no matter what. And we're not changing the mass of the object. OK. So stretching this spring further is not actually going to change the period. And so this period T two is gonna be the same as the period T one which is equal to three seconds. OK. So looking at our answer traces, we see that answer B three seconds is the correct answer. That's it for this one. Thanks everyone for watching. I hope this video helped.