Equations to be Remembered in respect of oscillations and simple harmonic motion were discussed in the post dated 17^{th} April 2008. This was followed by some multiple choice practice questions (with solution) and a free response practice question in the posts dated 22^{nd} April 2008 and 2^{nd} May 2008 respectively. A few multiple choice practice questions (with solution) on simple pendulum also were given later in the post dated 12^{th} September 2009. You can access all those posts by clicking on the label ‘oscillation’ or ‘simple harmonic motion’ below this post.

Today we will discuss a few more multiple choice practice questions (with solution) on simple harmonic motion. The following questions are meant for AP Physics B as well as AP Physics C aspirants:

**(1)** A girl is swinging on a swing in the sitting position. If she stands up, the the period of the swing will

(1) remain unchanged

(2) increase

(3) decrease

(4) become unpredictable

When the girl stands up, her centre of gravity is elevated and the effective length of the pendulum is *decreased*. The period of oscillation is therefore decreased. [Option (3)].

**(2)** The maximum velocity of a particle executing simple harmonic motion with an amplitude 6 cm, is 3.14 ms^{–1}. The period of oscillation is

(1) 120 s

(2) 12 s

(3) 1.2 s

(4) 0.12 s

The maximum velocity *v _{max} *is given by

*v _{max}*

_{ }=

*Aω*where

*A*is the amplitude and

*ω*is the angular frequency.

Therefore, *ω = v _{max}*

_{ }/

*A*

Or, 2π/*T **= v _{max}*

_{ }/

*A*from which

*T*= 2π

*A/*

*v*

_{max}Substituting known values, period *T* = 2π×6×10^{–2}/3.14 = 0.12 s

**(3) **One and of a light spiral spring is** **attached to a hook on the ceiling. A mass *m* kg hung on the spring stretches it by 10 cm. The mass is pulled down a little and released. The period oscillation of the system in seconds is (take *g = *10 ms^{–2})

(1) 2π*m/*5

(2) π*m/*5

(3) π*/*5

(4) 5π

The spring constant *k* is given by

*k =* *mg/x* = (*m*×10)/(0.1) = 100*m*

[Note that the extension *x* given in centimetre is converted into metre].

The period of oscillation (*T*) is given by

*T =*2π√(*m/k*) = 2π√(*m/*100*m*) = 2π/10 = π/5.

**(4)** You know that the period of oscillation *T *of a mass *m* attached to a light spring of force constant *k* is given by

*T =*2π√(*m/k*)

The period of oscillation of such a spring-mass system is found to be 2 s. If the period becomes 3 s when the mass is *increased by* 2 kg, what is the value of *m*?

(1) 0.8 kg

(2) 1 kg

(3) 1.2 kg

(4) 1.6 kg

Before adding the extra mass, we have

2 = 2π√(*m/k*)………….(i)

After adding the extra mass of 2 kg, we have

3= 2π√[(*m*+2)*/k*]………(ii)

Dividing Eq (i) by Eq (ii) we have

2/3 = √[*m/*(*m*+2)]

Squaring, 4/9 = *m/*(*m*+2).

This gives *m = *1.6 kg.

The following question is specifically for AP Physics C aspirants:

**(5)** A (1) A particle executing simple harmonic motion along the y-axis has zero displacement at time *t* = 0. The period of the motion is 1 s. After what time will its kinetic energy be 25% of the total energy?

(1) 1/12 s

(2) 1/6 s

(3) ¼ s

(4) 1 s

Since the displacement *y* is zero initially, the equation of the motion is

*y = A* sin *ωt* where *A* is the amplitude and *ω* is the angular frequency.

The velocity at the instant *t* is given by

*v* = d*y*/d*t* = *Aω* cos*ωt*

The kinetic energy at the instant *t* is given by

½ *mv*^{2} = ½ *mA*^{2}*ω*^{2}cos^{2}*ωt* where *m* is the mass of the particle*.*

The maximum kinetic energy which is *equal to the total energy* of the particle is ½ *mA*^{2}*ω*^{2}.

As the kinetic energy at time *t* is to be equal to 25% of the total energy, we have

½ *mA*^{2}*ω*^{2}cos^{2}*ωt* = ¼ ×½ *mA*^{2}*ω*^{2}

This gives cos*ωt* = ½ from which *ωt = *π/3.* *

Or, 2π*t*/*T **= *π/3

Since *T = *1 s, we get *t *= 1/6 s [Option (2)].

You will find additional multiple choice questions in this section here.

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