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III. Long Answers Questions · Q6

Q.Explain the horizontal oscillations of a spring.

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Step 1. Setup. A block of mass mm is attached to a massless spring of stiffness constant kk, the other end fixed, resting on a smooth (frictionless) horizontal surface, with equilibrium position x0x_0.

Step 2. Restoring force. If the mass is displaced through a small displacement xx from x0x_0 and released, the stretched (or compressed) spring exerts a restoring force F=−kxF=-kx, proportional to the displacement.

Step 3. Equation of motion. Newton's second law gives mx¨=−kxm\ddot x=-kx, i.e. x¨=−(k/m)x\ddot x=-(k/m)x -- the SHM equation.

Step 4. Time period. Comparing gives ω=k/m\omega=\sqrt{k/m}, frequency f=12πk/mf=\dfrac{1}{2\pi}\sqrt{k/m}, and time period T=2πm/kT=2\pi\sqrt{m/k}. …

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