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Prove the equation a. V² = u² + 2as b. s = ut + 1/2 at² 2) Derived an equation T ∝ m²lg

Physics

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Answer

b

T=klgT = k \sqrt{\frac{l}{g}}

Step-by-step solution

13 steps

Here are the derivations for the given equations:

1) Prove the equation

Part a· 4 steps

V2=U2+2aSV^2 = U^2 + 2aS

  1. We start with the definitions of acceleration and displacement for constant acceleration

    v=u+at(∗)v = u + at \quad (*) s=u+v2t(∗∗)s = \frac{u+v}{2}t \quad (**)
  2. From equation (∗)(*), we can express time tt

    t=v−uat = \frac{v-u}{a}
  3. Substitute this expression for tt into equation (∗∗)(**)

    s=u+v2(v−ua)s = \frac{u+v}{2} \left(\frac{v-u}{a}\right)
  4. Simplify the expression

    s=(v+u)(v−u)2as = \frac{(v+u)(v-u)}{2a} s=v2−u22as = \frac{v^2 - u^2}{2a} 2as=v2−u22as = v^2 - u^2 V2=U2+2aSV^2 = U^2 + 2aS

    The equation is proven.

Part b

T=klgT = k \sqrt{\frac{l}{g}}

S=Ut+12aT2S = Ut + \frac{1}{2} aT^2

  1. We use the definitions of average velocity and displacement for constant acceleration

    v=u+at(∗)v = u + at \quad (*) s=u+v2t(∗∗)s = \frac{u+v}{2}t \quad (**)
  2. Substitute the expression for vv from equation (∗)(*) into equation (∗∗)(**)

    s=u+(u+at)2ts = \frac{u + (u+at)}{2}t
  3. Simplify the expression

    s=2u+at2ts = \frac{2u + at}{2}t s=(u+12at)ts = \left(u + \frac{1}{2}at\right)t
  4. Distribute tt to obtain the final form

    S=Ut+12aT2S = Ut + \frac{1}{2}aT^2

    The equation is proven.

    2) Derive an equation for the period TT using dimensional analysis.

    The question asks to "Derive an equation T∝m2lgT \propto m^2lg". However, this expression is dimensionally inconsistent for a period. Assuming the intent is to derive the formula for the period of a simple pendulum using dimensional analysis, where the period TT depends on mass mm, length ll, and acceleration due to gravity gg, the derivation is as follows:

  5. Assume the period TT is proportional to some powers of mass mm, length ll, and acceleration due to gravity gg.

    Let T=kmalbgcT = k m^a l^b g^c, where kk is a dimensionless constant and a,b,ca, b, c are exponents to be determined.
    The dimensions of the quantities are:

    [T]=T[T] = \text{T} [m]=M[m] = \text{M} [l]=L[l] = \text{L} [g]=LT−2[g] = \text{LT}^{-2}
  6. Substitute these dimensions into the proportionality

    T1=MaLb(LT−2)c\text{T}^1 = \text{M}^a \text{L}^b (\text{LT}^{-2})^c M0L0T1=MaLb+cT−2c\text{M}^0 \text{L}^0 \text{T}^1 = \text{M}^a \text{L}^{b+c} \text{T}^{-2c}
  7. Compare the powers of M, L, and T on both sides of the equation to form a system of linear equations

    For M: a=0a = 0
    For L: b+c=0b+c = 0
    For T: 1=−2c1 = -2c

  8. Solve the system of equations

    From the T equation, c=−12c = -\frac{1}{2}.
    Substitute cc into the L equation: b+(−12)=0  ⟹  b=12b + (-\frac{1}{2}) = 0 \implies b = \frac{1}{2}.
    So, the exponents are a=0a=0, b=12b=\frac{1}{2}, c=−12c=-\frac{1}{2}.

  9. Substitute these exponents back into the proportionality

    T=km0l1/2g−1/2T = k m^0 l^{1/2} g^{-1/2} T=klgT = k \sqrt{\frac{l}{g}}

    The derived equation for the period of a simple pendulum is T=klg\boxed{T = k \sqrt{\frac{l}{g}}}, where kk is a dimensionless constant (experimentally found to be 2π2\pi).

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