A simple pendulum performs simple harmonic motion about x = 0 with an amplitude 'a' and time period T. The speed of the pendulum at x = a/2 will be

Question

A simple pendulum performs simple harmonic motion about x = 0 with an amplitude 'a' and time period T. The speed of the pendulum at x = a/2 will be (A)  ( π a/ T)  (B)  ( 3 π2 a/ T)  (C)  ( π a √ 3 / T)  (D)  ( π a √ 3 / 2 T)

Tardigrade Admin · Accepted Answer

For simple harmonic motion, v=ω √a2-x2. When x=(a/2), When x=(a/2), v=ω √a2-(a2/4)=ω √(3/4) a2. As ω=(2 π/T), ∴ v=(2 π/T) ⋅ (√3/2) a ⇒ v=(π √3 a/T).

Q. A simple pendulum performs simple harmonic motion about $x = 0$ with an amplitude 'a' and time period $T$ . The speed of the pendulum at $x = a /2$ will be

$\frac{πa}{T}$

$\frac{3 π ^{2} a}{T}$

$\frac{πa 3}{T}$

$\frac{πa 3}{2 T}$

For simple harmonic motion,
$v = ω a^{2} - x^{2}$ .
When $x = \frac{a}{2}$ ,
When $x = \frac{a}{2},$
$v = ω a^{2} - \frac{a ^{2}}{4} = ω \frac{3}{4} a^{2}$ .
As $ω = \frac{2 π}{T},$
$∴ v = \frac{2 π}{T} \cdot \frac{3}{2} a$
$\Rightarrow v = \frac{π 3 a}{T}$ .

Q. A simple pendulum performs simple harmonic motion about x=0 with an amplitude 'a' and time period T. The speed of the pendulum at x=a/2 will be

Tπa​

T3π2a​

Tπa3​​

2Tπa3​​