Figure shows the kinetic energy K of a simple pendulum versus its angle θ from the vertical. The pendulum bob has mass 0.2 kg. The length of the pendulum in meter (g=10 m / s 2) is

Question

Figure shows the kinetic energy K of a simple pendulum versus its angle θ from the vertical. The pendulum bob has mass 0.2 kg. The length of the pendulum in meter (g=10 m / s 2) is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Given that from graph (1/2) m Vm2=15 × 10-3 Vm=√0.150 m / s A ω=√0.150 m / s L θm ⋅ √(g/L)=√0.150 m / s ⇒ √g L=(√0.150/100 × 10-3) ⇒ L=(0.150/0.1)=1.5 m

Q. Figure shows the kinetic energy $K$ of a simple pendulum versus its angle $θ$ from the vertical. The pendulum bob has mass $0.2 k g$ . The length of the pendulum in meter $(g = 10 m / s^{2})$ is___

Given that from graph
$\frac{1}{2} m V_{m}^{2} = 15 \times 1 0^{- 3}$
$V_{m} = 0.150 m / s$
$A ω = 0.150 m / s$
$L θ_{m} \cdot \frac{g}{L} = 0.150 m / s$
$\Rightarrow gL = \frac{0.150}{100 \times 1 0 ^{- 3}}$
$\Rightarrow L = \frac{0.150}{0.1} = 1.5 m$

Q. Figure shows the kinetic energy K of a simple pendulum versus its angle θ from the vertical. The pendulum bob has mass 0.2kg. The length of the pendulum in meter (g=10m/s2) is___

Given that from graph 21​mVm2​=15×10−3 Vm​=0.150​m/s Aω=0.150​m/s Lθm​⋅Lg​​=0.150​m/s ⇒gL​=100×10−30.150​​ ⇒L=0.10.150​=1.5m

Q. Figure shows the kinetic energy $K$ of a simple pendulum versus its angle $θ$ from the vertical. The pendulum bob has mass $0.2 k g$ . The length of the pendulum in meter $(g = 10 m / s^{2})$ is___

Given that from graph
$\frac{1}{2} m V_{m}^{2} = 15 \times 1 0^{- 3}$
$V_{m} = 0.150 m / s$
$A ω = 0.150 m / s$
$L θ_{m} \cdot \frac{g}{L} = 0.150 m / s$
$\Rightarrow gL = \frac{0.150}{100 \times 1 0 ^{- 3}}$
$\Rightarrow L = \frac{0.150}{0.1} = 1.5 m$