Question

The period of a simple pendulum is found to increase by 50% when the length of the pendulum is increased by 0.6 m. Calculate the initial length and the initial period of oscillation at a place where g = 9.8 m s⁻².

Answer 1

let Initial length of simple pendulum is l

Let us consider, initial time period is T.

so, T = 2π√{l/g} ...... (1)

final length of simple pendulum, L = l + 0.6 m

final time period , T' = T + 50% of T = T + T/2 = 3/2 T

so, T' = 2π√{L/g}

or, 3T/2 = 2π√{(l + 0.6)/g} ......(2)

from equations (1) and (2),

{3T/2}/T = 2π√{(l + 0.6)/g}/2π√{l/g}

or, 3/2 = √{l + 0.6)/l}

or, 9/4 = (l + 0.6)/l

or, 9l = 4l + 2.4

or, 5l = 2.4

or, l = 0.48m

hence, initial length of simple pendulum is 0.48m

Answer 2

Answer:

(a) sin\omega t-cos\omega tsinωt−cosωt

= \sqrt{2}\left(sin\omega t\frac{1}{\sqrt{2}} - cos\omega t\frac{1}{\sqrt{2}}\right)

2

(sinωt

2

1

−cosωt

2

1

)

= \sqrt{2}(sin\omega tcos\pi/4-cos\omega tsin\pi/4)

2

(sinωtcosπ/4−cosωtsinπ/4)

= \sqrt{2}sin(\omega t-pi/4)

2

sin(ωt−pi/4)

hence, it is simple harmonic motion. and its period = 2π/\omegaω

(b) sin³ωt = 1/3(3sinωt - sin3ωt) [ from trigonometric formula ]

each term here, sinωt and sin3ωt represent SHM. But sin³ωt is the result of superposition of two SHMs. Hence, it is only periodic not SHM. Its time period is 2π/ω.

(c) It can be seen that it represents an SHM with a time period of 2π/ω.

(d) It represents periodic motion but not SHM. Its time period is 2π/ω.

(e) An exponential function never repeats itself. Hence, it is a non-periodic motion.

(f) It clearly represents a non-periodic motion