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.

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## 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 = m_{1} = 100g = 0.1kg

Extension in a spring = x_{1 }= 4cm = 0.04m

time = T = 0.568 sec

mass of object =?

Calculation:

using Hook’s law

F = kx_{1}

here F = m_{1}g

m_{1}g = kx_{1} k = m_{1}g/x_{1}

= 0.1 x9.8 / 0.04

k = 24.5 N/m

as for mass spring system

T = 2p √m_{2}/k

T^{2} = 4p^{2} m_{2}/k

m_{2 } = kT^{2}/4p^{2} _{ } = 24.5 x (0.568)^{2}/4x (3.14)^{2}

m_{2 }= 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 x_{0}^{2 }mgh = ½ kx_{0}^{2}

4 x 9.8 x 8.0 = ½ x 1960 x x_{o}^{2 }31.36 = 980 x_{o}^{2 }x_{o}^{2 }= 31.36/980

= 0.032

= 0.178

x_{o } = 0.18 m

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