Consider the situation shown in figure. If the blocks are displaced slightly in opposite directions and released, they will execute simple harmonic motion. The time period is

Question

Consider the situation shown in figure. If the blocks are displaced slightly in opposite directions and released, they will execute simple harmonic motion. The time period is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/774efa8ed5a35e8c091880d2f9eac920-.png" /> (A) 2 π √(m/k) (B) 2 π √(m/2 k) (C) 2 π √(m/5 k) (D) 2 π √(2 m/3 k)

Tardigrade Admin · Accepted Answer

When two masses oscillate as shown, the effective mass is taken as μ (reduced mass) =(m1 m2/m1+m2) Here, μ=(2 m ⋅ m/2 m+m)=(2/3) m Time period, T=2 π √(μ/k)=2 π √(2 m/3 k)

Q. Consider the situation shown in figure. If the blocks are displaced slightly in opposite directions and released, they will execute simple harmonic motion. The time period is

$2 π \frac{m}{k}$

$2 π \frac{m}{2 k}$

$2 π \frac{m}{5 k}$

$2 π \frac{2 m}{3 k}$

When two masses oscillate as shown, the effective
mass is taken as $μ$ (reduced mass) $= \frac{m _{1} m _{2}}{m _{1} + m _{2}}$
Here, $μ = \frac{2 m \cdot m}{2 m + m} = \frac{2}{3} m$
Time period, $T = 2 π \frac{μ}{k} = 2 π \frac{2 m}{3 k}$

Q. Consider the situation shown in figure. If the blocks are displaced slightly in opposite directions and released, they will execute simple harmonic motion. The time period is

2πkm​​

2π2km​​

2π5km​​

2π3k2m​​

When two masses oscillate as shown, the effective mass is taken as μ (reduced mass) =m1​+m2​m1​m2​​ Here, μ=2m+m2m⋅m​=32​m Time period, T=2πkμ​​=2π3k2m​​

$2 π \frac{m}{k}$

$2 π \frac{m}{2 k}$

$2 π \frac{m}{5 k}$

$2 π \frac{2 m}{3 k}$

When two masses oscillate as shown, the effective
mass is taken as $μ$ (reduced mass) $= \frac{m _{1} m _{2}}{m _{1} + m _{2}}$
Here, $μ = \frac{2 m \cdot m}{2 m + m} = \frac{2}{3} m$
Time period, $T = 2 π \frac{μ}{k} = 2 π \frac{2 m}{3 k}$