Lecture 1 PHY 2 Summer 25 final term

Lecture 1 15-1 Simple Harmonic Motion : Periodic Motion : If the motion of a body is such that it crosses from the same direction a particular point in its path of motion at regular interval, then the motion is called periodic motion or harmonic motion Simple Harmonic Motion : If acceleration of a body executing periodic motion acts along a fixed point in its path of motion in a such a way that its magnitude from that point is proportional to its displacement, then the motion of the body is called Simple Harmonic Motion. Such motion is a sinusoidal function of time t . That is, it can be written as a sine or a cosine of time t. Here we arbitrarily choose the cosine function and write the displacement (or position) of the particle in Fig.15-1 as x (t) = xm cos (ωt + φ) ; (displacement) ……. (1) Chapter 15: Oscillations.

Frequency : The frequency f of the oscillation is the number of times per second that it completes a full oscillation (a cycle) and has the unit of hertz (abbreviated Hz), where 1 hertz = 1 Hz = 1 oscillation per second = 1 s – 1 ………. (2).

Time Period : The time for one full cycle is the period T of the oscillation , which is T = 𝟏 𝒇 ………. (3) The angular frequency ω = 2 π T = 2πf ……. (4) The SI unit of angular frequency is the radian per second..

The amplitudes are different, but the frequency and period are the same. The amplitudes are the same, but the frequencies and periods are different. This negative value shifts the cosine curve rightward. This zero gives a regular cosine curve..

The Velocity of SHM :To find the velocity v(t) as a function of time , let’s take a time derivative of the position function x(t) in Eq. 1 : v(t) = 𝒅 𝒅𝒕 x(t) = 𝒅 𝒅𝒕 [ xm cos (ωt + φ) ] v(t) = - ω xm sin (ωt + φ) (velocity) …….. (6) ➢The velocity depends on time because the sine function varies with time, between the values of +1 and -1. ➢The quantities in front of the sine function determine the extent of the variation in the velocity, between + ωxm and - ωxm . We say that ωxm is the velocity amplitude vm of the velocity variation. vm= ωxm . ➢When the particle is moving rightward through x = 0, its velocity is positive and the magnitude is at this greatest value. ➢When it is moving leftward through x = 0,its velocity is negative and the magnitude is again at this greatest value. ➢This variation with time (a negative sine function) is displayed in the graph of Fig. b for a phase constant of φ = 0, which corresponds to the cosine function for the displacement versus time shown in Fig. a. x (t) = xm cos (ωt + φ).

The Acceleration of SHM : It can be found by differentiating the velocity function of Eq. 6 with respect to time to get the acceleration function of the particle in simple harmonic motion: a(t) = 𝒅 𝒅𝒕 v(t) = 𝒅 𝒅𝒕 [- ω xm sin (ωt + φ) ] a(t) = - ω2 xm cos (ωt + φ) (acceleration) ….. (7) ➢The acceleration varies because the cosine function varies with time, between +1 and -1.The variation in the magnitude of the acceleration is set by the acceleration amplitude am, which is the product ω2 xm that multiplies the cosine function. am= ω2 xm ➢Figure c displays Eq. 7 for a phase constant φ = 0, consistent with Figs. a and b. Note that the acceleration magnitude is zero when the cosine is zero , which is when the particle is at x = 0. ➢And the acceleration magnitude is maximum when the cosine magnitude is maximum, which is when the particle is at an extreme point, where it has been slowed to a stop so that its motion can be reversed..

x (t) = xm cos (ωt + φ) ; (displacement) ……. (1) a(t) = - ω2 xm cos (ωt + φ) (acceleration) …….. (7) Comparing Eqs. 1 and 7 we see an extremely neat relationship: a(t) = - ω2 x (t) …….. (8) In SHM, the acceleration a is proportional to the displacement x but opposite in sign, and the two quantities are related by the square of the angular frequency ω..

Linear simple harmonic oscillator [undamped oscillator] :The force law for simple harmonic motion Let us assume that there is no friction. Using Eq 8 we can apply Newton’s second law to describe the force responsible for SHM: F = ma = m(- ω2 x )= - (m ω2)x ………… (9) The minus sign means that the direction of the force on the particle is opposite the direction of the displacement of the particle. That is , in SHM the force is a restoring force in the sense that it fights against the displacement , attempting to restore the particle to the center point at x = 0 . Now for a block on a spring as in Fig. we know from Hooke’s law, F = - kx …………. (10) for the force acting on the block. F = - kx - (m ω2)x = - kx k= m ω2.

3 : What is the maximum acceleration of a platform that oscillates at amplitude 2.20 cm and frequency 6.60 Hz? xm = 2.20 cm = 0.0220 m f = 6.60 Hz a(t) = - ω2 xm cos (ωt + φ) am= ω2 xm = (2πf )2 xm = 4π2( 6.60)2( 0.0220) = 37.8 m/s2 = 37.8 m/s-s x (t) = xm cos (ωt + φ) v(t) = - ω xm sin (ωt + φ).

13 : An oscillator consists of a block of mass 0.500 kg connected to a spring. When set into oscillation with amplitude 35.0 cm, the oscillator repeats its motion every 0.500 s. Find the (a) period, (b) frequency, (c) angular frequency, (d) spring constant, (e) maximum speed, and (f) magnitude of the maximum force on the block from the spring. Given: m =0.500 kg xm =35.0 cm = 0.35 m T = 0.500 s (a) T = 0.500 s (b) f = 1 𝑇 = 1 0.500 = 2.00 Hz [2 oscillations/s] (c) ω = 2πf = 2π(2.00) = 12.6 rad/s (d) ω = 𝑘 𝑚. k = m ω2 = (0.500)(12.6)2 = 79.0 N/m (e) v(t) = - ω xm sin (ωt + φ) vm = ωxm = (12.6)(0.350) = 4.40 m/s (f) F = - k x Fs = k xm = (79.0)(0.350) = 27.6 N Newton’s third law, Fs = Fm = 27.6 N.

A block whose mass m is 680 g is fastened to a spring whose spring constant k is 65 N/m. The block is pulled a distance x = 11 cm from its equilibrium position at x = 0 on a frictionless surface and released from rest at t = 0. (a) What are the angular frequency, the frequency, and the period of the resulting motion? (b) What is the amplitude of the oscillation? (c) What is the maximum speed vm of the oscillating block, and where is the block when it has this speed? (d) What is the magnitude am of the maximum acceleration of the block? (e) What is the phase constant φ for the motion? (f) What is the displacement function x(t) for the spring–block system? Sample Problem 15.01: Block–spring SHM, amplitude, acceleration, phase constant.