Physics · Oscillations · NEET
No. The formula is T = 2 pi root(m/k) for both horizontal and vertical springs. Gravity only shifts the mean position downward by an amount x0 = mg/k. Once the mass sits at this new equilibrium, it oscillates about that point with the same T = 2 pi root(m/k). So g decides where the block rests, not how fast it oscillates.
Use Hooke's law first: k = F/x, where x is the extension in metres. Example: a force of 10 N stretches the spring by 5 cm (0.05 m), so k = 10/0.05 = 200 N/m. Then put this k into T = 2 pi root(m/k). This is the exact two-step trick tested in NEET 2021.
Because T depends on root(m), doubling m multiplies T by root(2) (about 1.41 times), not by 2. To double the time period you must make the mass 4 times bigger, since T is proportional to root(m). Many students wrongly assume T doubles when m doubles.
No, it is the same: T = 2 pi root(m/k). The only difference is the equilibrium position. A horizontal spring rests at its natural length; a vertical spring rests stretched by mg/k due to gravity. Both give identical time periods for the same m and k.
Use T proportional to root(m). Write T1/T2 = root(m1/m2) and square both sides. In NEET 2016, T = 3 s for mass m and T = 5 s for mass (m+1). So (3/5)^2 = m/(m+1) gives 9/25 = m/(m+1), which solves to m = 9/16 kg.
No. For an ideal spring obeying Hooke's law, T = 2 pi root(m/k) is independent of amplitude. Whether you pull the block 2 cm or 5 cm, the time for one full oscillation stays the same. This property is called isochronism and is true for all simple harmonic motion.
A spring is stretched by 5 cm under a force of 10 N. The time period of oscillation when a mass of 2 kg is suspended from it is
A body of mass m hangs from a spring of negligible mass. When pulled down and released it oscillates with a time period of 3 s. When the mass is increased by 1 kg, the time period becomes 5 s. The value of m (in kg) is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
T = 2 pi root(m/k), where m is the attached mass in kg and k is the spring constant in N/m. The frequency is f = 1/T = (1/2 pi) root(k/m).
Gravity is a constant force that only shifts the equilibrium position down by mg/k. It does not change the restoring behaviour about that new mean position, so T = 2 pi root(m/k) stays the same for vertical springs.
No. For an ideal spring, the time period is independent of amplitude. This is a key feature of simple harmonic motion.
A stiffer spring has a larger k. Since T is proportional to 1/root(k), a bigger k gives a smaller time period, so the block oscillates faster.
Since T is proportional to root(m), doubling the mass multiplies the time period by root(2), about 1.41 times, not by 2.
