Oscillations Question 1
Question 1 - 24 January - Shift 1
A block of mass $2 kg$ is attached with two identical springs of spring constant $20 N / m$ each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is $\frac{\pi}{\sqrt{x}}$ in SI unit. The value of $x$ is
m $2 Kg \ldots$
Show Answer
Answer: (5)
Solution:
Formula: Time period of spring mass system
$F=-2 kx, a=-\frac{2 kx}{m}, \omega=\sqrt{\frac{2 k}{m}}=\sqrt{\frac{2 \times 20}{2}}$
$=\sqrt{20} rad / s$
$T=\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{20}}=\frac{\pi}{\sqrt{5}}$
$x=5$
