FSC 1st Year Physics Chapter 7 Exercise & Numericals Notes PDF

In this post, I am sharing FSC 1st Year Physics Chapter 7 Solved Exercise & Numericals Notes PDF for the students of Intermediate Part 1. This is the 7th chapter from the First Year Physics book and its name is Oscillations. Students can download the 11th class physics chapter 7 Oscillations Solved Exercise & Numericals Notes PDF on their laptops and mobiles. These Physics Notes for Class 11 are for all the boards working under Punjab Board like Gujranwala Board, Lahore Board, Faisalabad Board, Multan Board, Rawalpindi Board, Sargodha Board, DG Kahn Board, and Sahiwal Board. Complete FSC First Year Physics Notes PDF are already available at ilmihub.com.

FS C Part 1 Physics Chapter 7 Oscillations Solved Exercise & Numericals Notes PDF Download

A 100-gram body hung on a spring clougates the spring by 4.0cm. When a certain object is hung on the spring and set vibrating, its period is 0.568s. What is the mass of the object pulling the spring?

Data

                                Mass of a body = m₁ = 100g = 0.1kg
                                Extension in a spring = x₁ = 4cm = 0.04m
                                time = T = 0.568 sec
                                mass of object =?
Calculation:
                using Hook’s law

                                F = kx₁
                here      F = m₁g
                                m₁g = kx₁
                                k = m₁g/x­₁
                                   = 0.1 x9.8 / 0.04
                                k = 24.5 N/m
as for mass spring system
                T = 2p √m₂/k
             T² = 4p² m­₂/k
                m₂  = kT²/4p²
                 = 24.5 x (0.568)²/4x (3.14)²
                m₂ = 0.20kg ans

A block of mass 4.0 kg dropped from a height of 0.80 m onto a spring of spring constant k = 1960 Nm^-1.
find the maximum distance through which the spring will be compressed

Data:     mass of block = m = 4kg
                height = h = 0.80m
                spring constant = k = 1960 N/m
                maximum distance =x = ?
calculation
                when the block is on the spring, the spring will compress due to the potential energy of the block
therefore,
                P.E = ½ k x₀²
                        mgh = ½ kx₀²
                4 x 9.8 x 8.0 = ½ x 1960 x x_o²
                        31.36 = 980 x_o²
                        x_o² = 31.36/980
                      = 0.032
                      = 0.178
                x_o  = 0.18 m

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