Question
A particle moves with simple harmonic motion in a straight line. In first $$\tau s,$$ after starting from rest it travels a distance $$a,$$ and in next $$\tau s$$ it travels $$2a,$$ in same direction, then:
A.
amplitude of motion is $$3a$$
B.
time period of oscillations is 8$$\tau $$
C.
amplitude of motion is $$4a$$
D.
time period of oscillations is 6$$\tau $$
Answer :
time period of oscillations is 6$$\tau $$
Solution :
In simple harmonic motion, starting from rest,
At $$t = 0,x = A$$
$$x = A\cos \omega t\,......\left( {\text{i}} \right)$$
When $$t = \tau ,x = A - a$$
When $$t = 2\tau ,x = A - 3a$$
From equation (i)
$$\eqalign{
& A - a = A\cos \omega \tau \,......\left( {{\text{ii}}} \right) \cr
& A - 3a = A\cos 2\omega \tau \,......\left( {{\text{iii}}} \right) \cr
& \cos 2\omega \tau = 2{\cos ^2}\omega \tau - 1......\left( {{\text{iv}}} \right) \cr} $$
From equation (ii), (iii) and (iv)
$$\eqalign{
& \frac{{A - 3a}}{A} = 2{\left( {\frac{{A - a}}{A}} \right)^2} - 1 \cr
& \Rightarrow \frac{{A - 3a}}{A} = \frac{{2{A^2} + 2{a^2} - 4Aa - {A^2}}}{{{A^2}}} \cr
& \Rightarrow {A^2} - 3aA = {A^2} + 2{a^2} - 4Aa \cr
& \Rightarrow 2{a^2} = aA \Rightarrow A = 2a \cr
& \Rightarrow \frac{a}{A} = \frac{1}{2} \cr
& {\text{Now,}}\,A - a = A\cos \omega \tau \cr
& \Rightarrow \cos \omega \tau = \frac{{A - a}}{A} \Rightarrow \cos \omega \tau = \frac{1}{2} \cr
& {\text{or,}}\,\frac{{2\pi }}{T}\tau = \frac{\pi }{3} \Rightarrow T = 6\tau \cr} $$