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प्रश्न
In following figure k = 100 N/m M = 1 kg and F = 10 N.
- Find the compression of the spring in the equilibrium position.
- A sharp blow by some external agent imparts a speed of 2 m/s to the block towards left. Find the sum of the potential energy of the spring and the kinetic energy of the block at this instant.
- Find the time period of the resulting simple harmonic motion.
- Find the amplitude.
- Write the potential energy of the spring when the block is at the left extreme.
- Write the potential energy of the spring when the block is at the right extreme.
The answer of b, e and f are different. Explain why this does not violate the principle of conservation of energy.
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उत्तर
It is given that :
Spring constant, k = 100 N/m,
Mass of the block, M = 1 kg
Force, F = 10 N
(a) In the equilibrium position,
F = kx
where x is the compression of the spring, and
k is the spring constant.
∴ `x = F/k = 10/100`
= 0.1 m = 10 cm
(b) The blow imparts a speed of 2 ms-1 to the block, towards left.
Potential energy of spring, U = `1/2 kx^2`
Kinetic energy,
`K = 1/2 mv^2`
(c) Time period (T) is given by,
`T = 2pisqrt(M/k)`
`2pi sqrt(1/100) = pi/5` s
(d) Let A be the amplitude.
Amplitude is the distance between the mean and the extreme position.
At the extreme position, the compression of the spring will be (A + x).
As the total energy in S.H.M. remains constant, we can write:
`1/2k(A + x )^2 = 1/2kx^2 + 1/2Mv^2 + FA`
= 2.5 + 10A
∴ 50(A + 0.1)2 = 2.5 + 10A
⇒ 50A2 + 0.5 + 10A = 2.5 + 10A
⇒ 50A2 = 2
⇒ `A^2 = 2/50 = 4/100`
⇒ `A = 2/10 m = 0.2 m = 20 cm`
(e) Potential Energy at the left extreme will be,
`P.E. = 1/2k(A + x)^2`
= `1/2 xx 100 xx (0.1 + 0.2)^2`
= 50 × (0.09) = 4.5 J
(f) Potential Energy at the right extreme is calculated as:
Distance between the two extremes = 2A
`P.E. = 1/2k(A + x)^2 - F(2A)`
= 4.5 − 10(0.4) = 0.5 J
As the work is done by the external force of 10 N, different values of options (b), (e), and (f) do not violate the law of conservation of energy.
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