If a particle undergoes shm with amplitude 0.10 m , what is the total distance it travels in one period?

If the amplitude is equal to 0.25m, then the distance travelled above and below the equilibrium is also equals to 0.25m. it means if it starts above the equilibrium, one period will be completed when it returns back to its original position so, it travels 0.25 below, 0.25m back, 0.25m down and o.25 m when it comes back to its original position, so the total distance we get is by adding 0.25m 4 times that is 1m.

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).