# Class 11 Physics Oscillations: NCERT Solutions for Question 10

This page focuses on the detailed Oscillations question answers for Class 11 Physics Oscillations, addressing the question: 'In Exercise 14.9, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is (a) at the mean position, (b) at the maximum stretched position, and (c) at the maximum compressed position. In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?'. The solution provides a thorough breakdown of the question, highlighting key concepts and approaches to arrive at the correct answer. This easy-to-understand explanation will help students develop better problem-solving skills, reinforcing their understanding of the chapter and aiding in exam preparation.
Question 10

## In Exercise 14.9, let us take the position of mass when the spring is unstreched as x = 0, and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is(a) at the mean position,(b) at the maximum stretched position, and(c) at the maximum compressed position.In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

Given,

Spring constant, k = 1200 N/m

Mass, m = 6 kg

Displacement, A = 4.0 cm = 0.04 cm

ω = 14.14 s-1

( i )

Since time is measured from mean position,

x = A sin ω t

x = 4 sin 14.14t

( ii ) At the maximum stretched position, the mass has an initial phase of π/2 rad.

Then, x = A sin( ωt + π/2 ) = A cos ωt

= 4 cos 14.14t

( iii ) At the maximum compressed position, the mass is at its leftmost position with an initial phase of 3π/2 rad.

Then, x = A sin( ωt + 3π/2 )

= -4 cos14.14 t