04 al phy q5(shm)

wing

2010-12-31 7:28 pm

Figure (1) shows a bolock of mass m suspended by a light spring .The extension of the spring is e .the spring is then cut into two identical parts and connected in parallel as shown is Figutr (2).If a block of mass 2 m is suspended by this set-up as shown ,What is the exension of each spring?
A.2e
B.E
C.e/2
D:e/4
The answer is C
Would you please explain briefly why the answer is C?

回答 (1)

天同

2010-12-31 7:56 pm

✔ 最佳答案

An elastic spring obeys Hooke' Law, T = k.x
where T is the spring tension, k is the spring constant and x is the extension

But k = EA/L, where E is the Modulus of Elasticity of the spring (which depends only on the spring material), L is the natural length of the spring, and A is the cross-sectional area of the spring

When the spring is cut into two, the new natural length L' become L/2, hence the new spring constant k' = EA/L' = 2EA/L = 2k. That is, each spring will now has a sping constant two times that before the spring is cut.

Let e' be the new extension, we have,
2mg = 2 x k'e'
i.e. 2mg = 2 x (2k).e'
e' = mg/2k

But before the spring was cut, we have mg = ke, or e = mg/k
hence, e' = e/2

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