Question

Different mass crates are placed on top of springs of uncompressed length L0 and stiffness k. (Figure 1) The crates are released and the springs compress to a length L before bringing the crates back up to their original positions.

Rank the time required for the crates to return to their initial positions from largest to smallest.

Rank from largest to smallest. To rank items as equivalent, overlap them.

ResetHelp 5N/mk 10 N/m L 10 cm L=5cm L 10 cm L=5cm Lo = 15 cm | | Lo = 15 cm | | Lo 20 cm | | Lo 20 cm L 5 cm Lo 10 cm | | Lo = 10 cm L=5cm largest smallest The correct rankina cannot be determined

Answer 1

As time period is the time taken for one complete oscillation, so, time taken by each crate to come back toits original posotion is nothing but the time period of the spring mass system, which is given as,    $\dpi{200} T$$\dpi{200} =$$\dpi{200} 2\pi$$\dpi{200} \sqrt$m/k

Now, at equillibrium,

Spring force upward = Weight of the mass down

Or, kx = mg

Or, m/k = x/g

But, equillibrium displacement, x = 1/2(Lo - L)

Where, Lo - L is the maximum distance by which the crane will fall before moving back upward, as the crane will oscillate by equal amount above and below the equillibrium position.

Hence, m/k = 1/2g(Lo - L)

Therefore, time period, T = 2$\dpi{200} \pi$$\dpi{200} \sqrt$1/2g(Lo - L)

So, out of the given cranes, the one with lowest value of (Lo - L) will take largest time and with highest value of (Lo - L) will take smallest time.

I will name the given cranes as 1 to 6 as given in the order in the question and calculate the value of (Lo - L) one by one and then will rank from largest to smallest as below

1--> (15-10) = 5 cm

2--> (15-5) = 10 cm

3--> (20-10) = 10 cm

4--> (20-5) = 15 cm

5--> (10-5) = 5 cm

6--> (10-5) = 5 cm

Thus, the rankings are, 1= 5 = 6 > 2 =3 > 4