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In following figure k = 100 N/m M = 1 kg and F = 10 N. (a) Find the compression of the spring in the equilibrium position. (b) A sharp blow by some external agent imparts a speed of 2 m/s to the block

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प्रश्न

In following figure k = 100 N/m M = 1 kg and F = 10 N. 

  1. Find the compression of the spring in the equilibrium position. 
  2. 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. 
  3. Find the time period of the resulting simple harmonic motion. 
  4. Find the amplitude. 
  5. Write the potential energy of the spring when the block is at the left extreme. 
  6. 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.

संख्यात्मक
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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.

shaalaa.com
Energy in Simple Harmonic Motion
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 12: Simple Harmonics Motion - Exercise [पृष्ठ २५३]

APPEARS IN

एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 12 Simple Harmonics Motion
Exercise | Q 16 | पृष्ठ २५३

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