Answer the following questions:(a) Time period of a particle in SHM depends on the force constant $k$ and mass $m$ of the particle: $\mathbf{T}==2 \pi(\sqrt{m} / \sqrt{k})$. A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?(b) The motion of a simple pendulum is approximately simple harmonic for small-angle oscillations. For larger angles of oscillation, a more involved analysis shows that $\mathbf{T}$ is greater than $2 \pi(\sqrt{I} / \sqrt{g}) .$ Think of a qualitative argument to appreciate this result.

Home » Answer the following questions:(a) Time period of a particle in SHM depends on the force constant and mass of the particle: . A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?(b) The motion of a simple pendulum is approximately simple harmonic for small-angle oscillations. For larger angles of oscillation, a more involved analysis shows that is greater than Think of a qualitative argument to appreciate this result.

Answer the following questions:
(a) Time period of a particle in SHM depends on the force constant $k$ and mass $m$ of the particle: $\mathbf{T}==2 \pi(\sqrt{m} / \sqrt{k})$ . A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
(b) The motion of a simple pendulum is approximately simple harmonic for small-angle oscillations. For larger angles of oscillation, a more involved analysis shows that $\mathbf{T}$ is greater than $2 \pi(\sqrt{I} / \sqrt{g}) .$ Think of a qualitative argument to appreciate this result.

CBSE | Class 11 | NCERT | Oscillations | Physics

Answer the following questions:
(a) Time period of a particle in SHM depends on the force constant $k$ and mass $m$ of the particle: $\mathbf{T}==2 \pi(\sqrt{m} / \sqrt{k})$ . A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
(b) The motion of a simple pendulum is approximately simple harmonic for small-angle oscillations. For larger angles of oscillation, a more involved analysis shows that $\mathbf{T}$ is greater than $2 \pi(\sqrt{I} / \sqrt{g}) .$ Think of a qualitative argument to appreciate this result.

(a) The spring constant $k$ is proportional to the mass in the case of a simple pendulum. The numerator ( $m$ ) and denominator ( $d$ ) will cancel each other out. As a result, the simple pendulum’s time period is independent of the bob’s mass.