Answer:
The right answer is:
(a) 63.83 kg
(b) 0.725 m/s
Explanation:
The given query seems to be incomplete. Below is the attachment of the full question is attached.
The given values are:
T = 3 sec
k = 280 N/m
(a)
The mass of the string will be:
⇒ [tex]T=2 \pi\sqrt{\frac{m}{k} }[/tex]
or,
⇒ [tex]m=\frac{k T^2}{4 \pi^2}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{280\times (3)^2}{4 \pi^2}[/tex]
⇒ [tex]=\frac{280\times 9}{4\times (3.14)^2}[/tex]
⇒ [tex]=68.83 \ kg[/tex]
(b)
The speed of the string will be:
⇒ [tex]\frac{1}{2}k(0.4)^2=\frac{1}{2}k(0.2)^2+\frac{1}{2}mv^2[/tex]
then,
⇒ [tex]v=\sqrt{\frac{k((0.4)^2-(0.2)^2)}{m} }[/tex]
On substituting the values, we get
⇒ [tex]=\sqrt{\frac{280\times ((0.4)^2-(0.2)^2)}{63.83} }[/tex]
⇒ [tex]=\sqrt{\frac{280(0.16-0.04)}{63.83} }[/tex]
⇒ [tex]=\sqrt{\frac{280\times 0.12}{63.83} }[/tex]
⇒ [tex]=0.725 \ m/s[/tex]
