A simple pendulum of time period 1s and length l is hung from a fixed support at O, such that the bob is at a distance H vertically above A on the ground (Fig) The amplitude is θ o . The string snaps at θ = θ₀ /2. Find the time taken by the bob to hit the ground. Also find distance from A where bob hits the ground. Assume θ₀ to be small so that sin _o ₀ and cos ₀

2 Views|Posted 7 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Payal Gupta | Contributor-Level 10

7 months ago

This is a long answer type question as classified in NCERT Exemplar

Let us assume t=0 when $θ = θ$ _o then $θ$ = $θ$ ₀coswt

Given a seconds pendulum w=2 $π$

$θ$ = $θ_{o} c o s 2 π t$

At time t, let $θ = θ_{o} / 2$

Cos2 $π t_{1} = \frac{1}{2}$ , t₁=1/6

$\frac{d θ}{d t}$ =-( $θ_{o} 2 π$ )sin2 $π t$

t=t₁=1/6

d $\frac{θ}{d t}$ =- $θ_{o} 2 π$ sin $\frac{2 π}{6} = - \sqrt[]{3} π θ_{o}$

the linear velocity is u=- $\sqrt[]{3} π θ_{o} l$

the vertical component is U_y= $- \sqrt[]{3} π l s i n (\frac{θ_{o}}{2})$

the horizontal component U

Upvote

Similar Questions for you

If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is:

Vishal Baghel

$T = 2 π \sqrt[]{\frac{l}{g}}$

$\Rightarrow 2 = 2 π \sqrt[]{\frac{2}{g}}$

$\Rightarrow g = 2 π^{2} m / s^{2}$

In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. New amplitude of oscillation will be:

Vishal Baghel

Velocity of block in equilibrium, in first case,

$v = A ω = A . \sqrt[]{\frac{k}{M}}$

Velocity of block in equilibrium, is second case,

$v' = A' ω' = A' \sqrt[]{\frac{k}{M + m}}$

From conservation of momentum,

Mv = (M + m) v’

$\Rightarrow M A \sqrt[]{\frac{k}{M}} = (M + m) A' \sqrt[]{\frac{k}{M + m}} \Rightarrow A' = A \sqrt[]{\frac{M}{M + m}}$

The fundamental frequency of an organ pipe open at one end is 300 Hz. The frequency of 3rd overtone of n Find n.×this organ pipe is 100 Hz

Vishal Baghel

f? = 300 Hz
3rd overtone = 7f? = 2100 Hz

Consider two identical springs each of spring constant k and negligible mass compared to the mass M as shown. Fig. 1 shows one of them and Fig. 2 shows their series combination. The ratios of time period of oscillation of the two SHM is T_b / T_a = √x, where value of x is ................... (Round off to the Nearest Integer)

Vishal Baghel

Kindly consider the following figure

For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal?


Vishal Baghel

K = U

½ mω² (A² - x²) = ½ mω²x²

A² - x² = x²

A² = 2x²

x = ± A/√2

Taking an Exam? Selecting a College?

Get authentic answers from experts, students and alumni that you won't find anywhere else.

On Shiksha, get access to

66K

Colleges

1.2K

Exams

6.9L

Reviews

1.8M

Answers

Learn more about...

physics ncert solutions class 11th 2023

View Exam Details

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

Ask Current Students, Alumni & our Experts

Quick Actions

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Have a question related to your career & education?

See what others like you are asking & answering