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If a mass m is attached to an ideal massless spring and has a period of t, then the period of the system when the mass is 2m is
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Calculation:
Step-1:
It is given that a mass m is attached to an ideal massless spring and the period of the system is t. It is required to find the period when the mass is doubled.
The time it takes an object to complete one oscillation and return to its initial position is measured in terms of a period, or T.
It is known that the period is calculated as,

Here m is the mass of the object, and k is the spring constant.
Step-2:
Thus the period of the system with the first mass is,

The period of the system with the second mass is,

Then the period of the system with the second mass is
times more than the period of the system with the first mass.
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This question involves the concepts of work done and elastic potential energy.
The work done by the spring when its extension changes from X to X/4 is "0.56 E".
The work done by the spring is equal to the elastic potential energy stored in the spring, which is given as follows:

where,
W = work done = ?
E = elastic potential energy
k = spring constant
X = extension
Now, the spring moves to an extension of X/4, so the change in extension will be:

Hence, the work done will become:

<u>W = 0.56 E</u>
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Learn more about Elastic Potential Energy here:
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Attached picture shows Elastic Potential Energy.
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