Answer the following questions:
$(a)$ Time period of a particle in $SHM$ depends on the force constant $k$ and mass $m$ of the particle:
$T=2 \pi \sqrt{\frac{m}{k}}$. A stmple pendulum executes $SHM$ approximately. Why then is the time pertodof.anondwers period of a pendulum independent of the mass of the pendulum?
$(b)$ The motion of a simple pendulum is approximately stmple harmonte for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that $T$ is greater than $2 \pi \sqrt{\frac{l}{g}} .$ Think of a qualitative argument to appreciate this result.
$(c)$ A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?
$(d)$ What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely failing under gravity?
The time period of a simple pendulum, $T=2 \pi \sqrt{\frac{m}{k}}$
For a simple pendulum, $k$ is expressed in terms of mass $m,$ as:
$k{\propto} m$
$\frac{m}{k}=$ Constant
Hence, the time period $T$, of a simple pendulum is independent of the mass of the bob.
In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as:
$F=-m g \sin \theta$
Where,
$F=$ Restoring force
$m=$ Mass of the bob
$g=$ Acceleration due to gravity
$\theta=$ Angle of displacement
For small $\theta, \sin \theta=\theta$
For large $\theta, \sin \theta$ is greater than $\theta$
This decreases the effective value of $g$.
Hence, the time period increases as:
$T=2 \pi \sqrt{\frac{l}{g}}$
Where, $l$ is the length of the simple pendulum
The time shown by the wristwatch of a man falling from the top of a tower is not affected by the fall. since a wristwatch does not work on the principle of a simple pendulum, it is not affected by the acceleration due to gravity during free fall. Its working depends on spring action.
When a simple pendulum mounted in a cabin falls freely under gravity, its acceleration is zero. Hence the frequency of oscillation of this simple pendulum is zero.
If the length of a pendulum is made $9$ times and mass of the bob is made $4$ times then the value of time period becomes
A simple pendulum of length $l$ is made to oscillate with an amplitude of $45$ degrees. The acceleration due to gravity is $g$. Let $T_0=2 \pi \sqrt{l / g}$. The time period of oscillation of this pendulum will be
For a simple pendulum the graph between $L$ and $T$ will be.
What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of $10cm$ ......... $m/s$ (Take $g = 9.8\,m/{s^2})$
Match $List - I$ with $List - II$
Choose the correct answer from the options given below