Let Us Take the Position of Mass When the Spring is Unstretched As X = 0, and the Direction from Left to Right as the Positive Direction of the X-axis. Give X As a Function of Time T for the Oscillating Mass If at the Moment We Start the Stopwatch (T - Physics

Advertisements
Advertisements

let us take the position of mass when the spring is unstretched as x = 0, and the direction from left to right as the positive direction of the x-axis. Give as a function of time t for the oscillating mass if at the moment we start the stopwatch (= 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?

Advertisements

Solution 1

(a) x = 2sin 20t

(b) x = 2cos 20t

(c) x = –2cos 20t

The functions have the same frequency and amplitude, but different initial phases.

Distance travelled by the mass sideways, A = 2.0 cm

Force constant of the spring, k = 1200 N m–1

Mass, m = 3 kg

Angular frequency of oscillation:

`omega =  sqrt(k/m)`

`= sqrt(1200/3)= sqrt400  = 20 rad s^(-1)`               

a) When the mass is at the mean position, initial phase is 0.

Displacement, x = Asin ωt

= 2sin 20t

b) At the maximum stretched position, the mass is toward the extreme right. Hence, the initial phase is `pi/2`

Displacement , `x = Asin(omegat + pi/2)`

`= 2sin (20t + pi/2)`

= 2cos 20t

(c) At the maximum compressed position, the mass is toward the extreme left. Hence, the initial phase is `(3pi)/2`

Displacement, `x = Asin(omegat + 3pi/2)`

`= 2sin (20t + 3pi/2) = - 2 cos 20 t`

The functions have the same frequency  (`20/(2pi) Hz`)  and amplitude (2 cm), but different initial phases `(0, pi/2, (3pi)/2)`

Solution 2

a =2 cm, omega = `sqrt(k/m) = sqrt(1200/3) s^(-1)= 20s^(-1)`

a) Since time s measured from mean position

b) At the maximum stretched position, tyhe body is at the extreme right position. The initial phase is `pi/2`

`:. x = a sin (omegat + pi/2) = a cos omegat =  2 cos  20 t`

c) At the maximum compressed position, the body is at the extreme left position. The initial phase is `(3pi)/2`

`:. x = a sin (omegat  + (3pi)/2) = - a cosomegat = - 2 cos 20t`

  Is there an error in this question or solution?
Chapter 14: Oscillations - Exercises [Page 359]

APPEARS IN

NCERT Physics Class 11
Chapter 14 Oscillations
Exercises | Q 10 | Page 359

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

The period of a conical pendulum in terms of its length (l), semi-vertical angle (θ) and acceleration due to gravity (g) is:


If the metal bob of a simple pendulum is replaced by a wooden bob of the same size, then its time period will.....................

  1. increase
  2. remain same
  3. decrease
  4. first increase and then decrease.

When the length of a simple pendulum is decreased by 20 cm, the period changes by 10%. Find the original length of the pendulum.


The phase difference between displacement and acceleration of a particle performing S.H.M. is _______.

(A) `pi/2rad`

(B) π rad

(C) 2π rad

(D)`(3pi)/2rad`


A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.

Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass.


Answer the following questions:

A time period of a particle in SHM depends on the force constant and mass of the particle: `T = 2pi sqrt(m/k)` A simple pendulum executes SHM approximately. Why then is the time 

 


Answer the following questions:

The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that is greater than `2pisqrt(1/g)`  Think of a qualitative argument to appreciate this result.


The cylindrical piece of the cork of density of base area and height floats in a liquid of density `rho_1`. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period

`T = 2pi sqrt((hrho)/(rho_1g)` 

where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).


A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance x0 and pushed towards the centre with a velocity v0 at time = 0. Determine the amplitude of the resulting oscillations in terms of the parameters ω, x0 and v0. [Hint: Start with the equation acos (ωt) and note that the initial velocity is negative.]


A clock regulated by seconds pendulum, keeps correct time. During summer, length of pendulum increases to 1.005 m. How much will the clock gain or loose in one day?

(g = 9.8 m/s2 and π = 3.142)


Define practical simple pendulum


Show that motion of bob of the pendulum with small amplitude is linear S.H.M. Hence obtain an expression for its period. What are the factors on which its period depends?


Show that, under certain conditions, simple pendulum performs the linear simple harmonic motion.


If the particle starts its motion from mean position, the phase difference between displacement and acceleration is ______.


A simple pendulum has a time period of T1 when on the earth's surface and T2 when taken to a height R above the earth's surface, where R is the radius of the earth. The value of `"T"_2 // "T"_1` is ______. 


If the maximum velocity and acceleration of a particle executing SHM are equal in magnitude, the time period will be ______.


The period of oscillation of a simple pendulum of constant length at the surface of the earth is T. Its time period inside mine will be ______.


Which of the following statements is/are true for a simple harmonic oscillator?

  1. Force acting is directly proportional to displacement from the mean position and opposite to it.
  2. Motion is periodic.
  3. Acceleration of the oscillator is constant.
  4. The velocity is periodic.

Find the time period of mass M when displaced from its equilibrium position and then released for the system shown in figure.


A body of mass m is situated in a potential field U(x) = U0 (1 – cos αx) when U0 and α are constants. Find the time period of small oscillations.


A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period. `T = 2πsqrt(m/(Apg))` where m is mass of the body and ρ is density of the liquid.


A tunnel is dug through the centre of the Earth. Show that a body of mass ‘m’ when dropped from rest from one end of the tunnel will execute simple harmonic motion.


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. The new amplitude of oscillation will be:


A pendulum of mass m and length ℓ is suspended from the ceiling of a trolley which has a constant acceleration a in the horizontal direction as shown in the figure. Work done by the tension is ______.

(In the frame of the trolley)

 


A particle at the end of a spring executes simple harmonic motion with a period t1, while the corresponding period for another spring is t2. If the period of oscillation with the two springs in series is T, then ______.


Share
Notifications



      Forgot password?
Use app×