Class 11 Physics Oscillations Oscillations due to spring

Oscillations due to spring

Consider a block if it is pulled on one side and is released, and then it executes to and fro motionabout a mean position.

In the above image ablock, is on a frictionless surface when pulled or pushed and released, executes simple harmonic motion.

F (x) = –k x (expression for restoring force)

  • kis known as spring constantand its value is governed by the elastic properties of the spring. 
  • The above expression is same as the force law for SHM andtherefore the system executes a simple harmonicmotion. Therefore, 
  • ω = √k/m 
  • T = 2√m/k where T is the period.

Problem: - A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 s. What is the weight of the body?

Answer:

Maximum mass that the scale can read, M = 50 kg

Maximum displacement of the spring = Length of the scale, l = 20 cm = 0.2 m

Time period, T = 0.6 s

Maximum force exerted on the spring, F = Mg

where,g = acceleration due to gravity = 9.8 m/s2

F = 50 × 9.8 = 490 N

∴spring constant, k = F/l = 490/0.2 = 2450 N m-1.

Mass m, is suspended from the balance  

∴Weight of the body = mg = 22.36 × 9.8 = 219.167 N

Hence, the weight of the body is about 219 N.

 

Problem: - A 5 kg collar is attached to a spring of spring constant 500 N m–1. It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 10.0 cm and released. Calculate

(a) the period of oscillation,

(b) the maximum speed and

(c) maximum acceleration of the collar.

Answer:-

The period of oscillation as given by

T =2√m/k = 2 π√ (5.0 kg/500 Nm-1)

 = (2π/10)s

= 0.63 s

(b) The velocity of the collar executing SHM is

given by,

v (t) = –Aω sin (ωt + φ)

The maximum speed is given by,

vm = Aω

 = 0.1√ (k/m)

= 0.1√ (500Nm-1/5kg)

= 1 ms–1

and it occurs at x = 0

(c) The acceleration of the collar at thedisplacement x (t) from the equilibrium isgiven by,

a (t) = –ω2x (t)

 = – k/m x (t)

Therefore the maximum acceleration is,

 amax = ω2 A

 = (500 N m–1/5kg) x0.1m

= 10 m s–2and it occurs at the extreme positions.

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