Spring constant (force constant of a spring) is the force required for unit extension (or contraction) in a spring. Suppose a spring is cut into two pieces of equal length. Will the spring constant change? Don’t be doubtful. The spring constant will be doubled. If you cut a spring of spring constant ‘k’ into ‘n’ pieces of equal length, the spring constant of each piece will be nk.
Springs may appear in series and parallel combinations in certain questions, as for example, in problems involving the period of oscillation of a spring-mass system. If you have springs of constants k1, k2, k3….etc. in series, the net spring constant ‘k’ of the combination is given by the reciprocal relation, 1/k = 1/k1 + 1/k2 + 1/k3 + . . . . . . .etc.
If you have springs in parallel, the net spring constant ‘k’of the combination is given by k = k1+ k2+ k3+ . . . . .etc.
Often you will encounter questions involving the potential energy of a spring. You should remember that the potential energy of a spring of constant ‘k’, stretched (or contracted) through a distance ‘x’ is (½) kx2. Consider the following MCQ:
When a long spiral spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 6 cm, its potential energy will be
(a) 3U (b) 6U (c) 9U (d) U (e) 36U
Since the potential energy is ½ kx2, it follows that the P.E. is directly proportional to the square of the stretch (extension). The extension being 3 times, the P.E. must be 9 times. So, the correct option is (c).
Suppose the above question is asked in the following modified form:
When a long spiral spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 6 cm, the increase in its potential energy will be
(a) 4U (b) 6U (c) 9U (d) 8U (e) 36U
Since the difference between the initial and final potential energies is required in this problem, the answer is 9U-U = 8U.
Now, consider the following question:
The tension in a spring of spring constant k is T. The potential energy of the spring is
(a) T2/k2 (b) T2/k (c) 2T2/k (d) T2/2k (e) 2T2/k2
Potential energy of a spring, as you know, is (½)kx2. But k=T/x from which x=T/k. On substituting this value of x in the expression for potential energy, we obtain option (d) as the answer.
Now, consider the following M.C.Q. which appeared in IIT screening 2002question paper:
An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially un-stretched. Then the maximum extension in the spring is
(a) 4Mg/k (b) 2Mg/k (c)Mg/k (d) Mg/2k
When the mass is released, its gravitational potential energy is decreased and the elastic potential energy in the spring is increased by an equal amount. If ‘x’ is the maximum extension produced (when the mass reaches the lowest position), we have Mgx = (½) kx2, from which x = 2Mg/k. [Option (b)]
Springs may appear in series and parallel combinations in certain questions, as for example, in problems involving the period of oscillation of a spring-mass system. If you have springs of constants k1, k2, k3….etc. in series, the net spring constant ‘k’ of the combination is given by the reciprocal relation, 1/k = 1/k1 + 1/k2 + 1/k3 + . . . . . . .etc.
If you have springs in parallel, the net spring constant ‘k’of the combination is given by k = k1+ k2+ k3+ . . . . .etc.
Often you will encounter questions involving the potential energy of a spring. You should remember that the potential energy of a spring of constant ‘k’, stretched (or contracted) through a distance ‘x’ is (½) kx2. Consider the following MCQ:
When a long spiral spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 6 cm, its potential energy will be
(a) 3U (b) 6U (c) 9U (d) U (e) 36U
Since the potential energy is ½ kx2, it follows that the P.E. is directly proportional to the square of the stretch (extension). The extension being 3 times, the P.E. must be 9 times. So, the correct option is (c).
Suppose the above question is asked in the following modified form:
When a long spiral spring is stretched by 2 cm, its potential energy is U. If the spring is stretched by 6 cm, the increase in its potential energy will be
(a) 4U (b) 6U (c) 9U (d) 8U (e) 36U
Since the difference between the initial and final potential energies is required in this problem, the answer is 9U-U = 8U.
Now, consider the following question:
The tension in a spring of spring constant k is T. The potential energy of the spring is
(a) T2/k2 (b) T2/k (c) 2T2/k (d) T2/2k (e) 2T2/k2
Potential energy of a spring, as you know, is (½)kx2. But k=T/x from which x=T/k. On substituting this value of x in the expression for potential energy, we obtain option (d) as the answer.
Now, consider the following M.C.Q. which appeared in IIT screening 2002question paper:
An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially un-stretched. Then the maximum extension in the spring is
(a) 4Mg/k (b) 2Mg/k (c)Mg/k (d) Mg/2k
When the mass is released, its gravitational potential energy is decreased and the elastic potential energy in the spring is increased by an equal amount. If ‘x’ is the maximum extension produced (when the mass reaches the lowest position), we have Mgx = (½) kx2, from which x = 2Mg/k. [Option (b)]
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