A pendulum of  length 1cm hangs from an inclined wall. Suppose that this pendulum is released at an initial angle of  10∘ and it bounces off the wall elastically when it reaches an angle of  −5∘ as shown in the figure. (Take g=π2 m/s2) The period of this pendulum is (in second)

In  right half of oscillation of bob it oscillates 10∘ so amplitude is  θ0=10°In left half it undergoes only 5° due to the obstruction of wall. It travels only 5° that is θ02.Using the equation for angular SHM  θ=θ0sinωt ……( 1 )Time (t) to travel from 0 to  θ02,substituting θ=θ02 in equation ( 1 )θ02=θ0sinωt⇒ωt=30∘∵ω=2πT⇒t=T12Simple harmonic oscillator takes  T12 seconds. So total time period of the oscillation isT'=T4+T12+T4+T12=23Twhere T=2πlg=2s∴T'=23.2=43=1.33sTherefore, the correct answer is 1.33