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Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.
We discuss problems, which highlight certain aspects of the study leading to spring. The questions are categorized in terms of the characterizing features of the subject matter :
Problem 1 : A mass of 5 kg is suspended with two “mass-less” spring balances as shown in the figure. What would be the reading in each of the spring balances.
Solution : The spring balance is calibrated to measure the mass of the object suspended from it. In the stabilized condition, the spring force in the lower spring is equal to the weight of the object suspended from it. Since spring balances are "mass-less", the lower spring reads 5 kg.
The lower "spring balance" and the "object" together are attached to the lower end of the upper spring balance. Again, a total of 5 kg is suspended from the upper spring balance (note that springs have no weight). As such, the spring force in the upper spring is equal to the weight of the object suspended from it. The upper spring, therefore, also reads 5 kg.
Problem 2 : In the arrangement shown in the figure, the block is displaced by a distance “x” towards right and released. Find the acceleration of the block immediately after it is released.
Solution : When block is displaced towards right, the spring on left is stretched. The spring applies a force on the block towards left as spring tries to contract. The force is equal to the spring force and is given by :
The force is opposite to the assumed positive reference direction of displacement (towards right). On the other hand, the spring on right is compressed. As such, it also applies a force towards left as spring tries to expand.
The force is equal to the spring force and is given by :
Thus, we see that both springs apply force in the direction opposite to the reference direction. The net force on the block, at the time of extension “x” in the spring is given by :
The acceleration of the block, immediately after it is released, is :
The acceleration of the block is negative as all other quantities in the above expression are positive. It means that acceleration is in the opposite direction of the displacement. The magnitude of acceleration is :
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