A simple pendulum performs simple harmonic motion about x = 0 with an amplitude (A) and time period (T). The speed of the pendulum at $x = \frac{A}{2}$ 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 = \frac{A}{2}

will be

Toppr Admin · Accepted Answer

As simple pendulum performs simple harmonic motion-x2234-velocity-v-x3C9-x221A-a2-x2212-x2At- x-a2v-2-x3C0-T-x221A-a2-x2212-a2-2-2-x3C0-T-x221A-3a22-x3C0-a-x221A-3T

A simple pendulum performs simple harmonic motion about $x = 0$ with an amplitude $^{'} a^{'}$ and time period $^{'} T^{'}$ . The speed of the pendulum at $x = \frac{a}{2}$ will be:
$\frac{π a}{T}$
$\frac{3 π^{2} a}{T}$
$\frac{π a \sqrt{3}}{T}$
$\frac{π a \sqrt{3}}{2 T}$

As simple pendulum performs simple harmonic motion.
$∴ v e l o c i t y, v = ω \sqrt{a^{2} - x^{2}}$
At, $x = \frac{a}{2}$
$v = \frac{2 π}{T} \sqrt{a^{2} - {(\frac{a}{2})}^{2}} = \frac{2 π}{T} \frac{\sqrt{3 a^{2}}}{2} = \frac{π a \sqrt{3}}{T}$

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=a2 will be:πaT3π2aTπa√3Tπa√32T

As simple pendulum performs simple harmonic motion.∴velocity,v=ω√a2−x2At, x=a2v=2πT√a2−(a2)2=2πT√3a22=πa√3T

A simple pendulum performs simple harmonic motion about $x = 0$ with an amplitude $^{'} a^{'}$ and time period $^{'} T^{'}$ . The speed of the pendulum at $x = \frac{a}{2}$ will be:
$\frac{π a}{T}$
$\frac{3 π^{2} a}{T}$
$\frac{π a \sqrt{3}}{T}$
$\frac{π a \sqrt{3}}{2 T}$

As simple pendulum performs simple harmonic motion.
$∴ v e l o c i t y, v = ω \sqrt{a^{2} - x^{2}}$
At, $x = \frac{a}{2}$
$v = \frac{2 π}{T} \sqrt{a^{2} - {(\frac{a}{2})}^{2}} = \frac{2 π}{T} \frac{\sqrt{3 a^{2}}}{2} = \frac{π a \sqrt{3}}{T}$