A 100 g mass stretches a particular spring by 9.8 cm, when suspended vertically from it. How large a mass must be attached to the spring if the period of vibration is to be 6.28 s ?

Question

A 100 g mass stretches a particular spring by 9.8 cm, when suspended vertically from it. How large a mass must be attached to the spring if the period of vibration is to be 6.28 s ? (A) 1000 g (B) 105 g (C) 107 g (D) 104 g

Tardigrade Admin · Accepted Answer

At point of equilibrium k x=m g k × 9.8 × 10-3=100 × 10-3 × 9.8 k=100 × 10-1 k=10 N / m Period of vibration needed =6.28 s T=2 π √(m/k) 6.28=2 × 3.14 √(m/10) 1=(m/10) m=10 kg or 104 g

Q. A $100 g$ mass stretches a particular spring by $9.8 c m$ , when suspended vertically from it. How large a mass must be attached to the spring if the period of vibration is to be $6.28 s$ ?

$1000 g$

$1 0^{5} g$

$1 0^{7} g$

$1 0^{4} g$

At point of equilibrium $k x = m g$
$k \times 9.8 \times 1 0^{- 3} = 100 \times 1 0^{- 3} \times 9.8$
$k = 100 \times 1 0^{- 1}$
$k = 10 N / m$
Period of vibration needed $= 6.28 s$
$T = 2 π \frac{m}{k}$
$6.28 = 2 \times 3.14 \frac{m}{10}$
$1 = \frac{m}{10}$
$m = 10 k g$ or $1 0^{4} g$

Q. A 100g mass stretches a particular spring by 9.8cm, when suspended vertically from it. How large a mass must be attached to the spring if the period of vibration is to be 6.28s ?

1000g

105g

107g

104g