Login/Sign Up
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.
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.

(b) The expression for the restoring force acting on the bob of a basic pendulum is

$F=-m g \sin \theta$

$F=$ restoring force

$\mathrm{m}=$ mass of the bob

$\mathrm{g}=$ acceleration due to gravity

$\theta=$ angle of displacement

When $\theta$ is small, $\sin \theta \approx \theta$.

Then the expression for the time period of a simple pendulum is given by $\mathrm{T}=2 \pi(\sqrt{I} / \sqrt{g})$

$sin \theta < \theta$ when $\theta$ is huge. As a result, the equation above is invalid. In the time period $T$, there will be an increase.

i

More Material